Mahesh

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9b

I. In Exercises 1 to 7, find the equation of the line which satisfy the given conditions.

Question 1.
Write the equations for the X and Y – axes.
Solution:
The y-coordinate of every point on the X-axis is 0. Therefore, the equation of the X-axis is y = 0. The ‘x’ coordinate of every point on the Y-axis is zero.
Therefore, the equation of Y-axis is x = 0.

Question 2.
Find the equation of line passing through the point (-4, 3) with slope ½.
Solution:
We know that the equation of the line passing through the point (x₀, y₀) whose slope is m is (y – y₀) = m(x – x₀).
Thus, the equation of the line passing through the point (- 4, 3), whose slope is ½
(y – 3) = ½(x + 4)
⇒ 2(y -3) = x + 4
⇒ 2y – 6 = x + 4
⇒ x – 2y + 10 = 0

Question 3.
Find the equation of the line passing through (0, 0) with slope m.
Solution:
We know that the equation of the line passing through the point (x₀, y₀) whose slope’m’ is (y – y₀) = m(x – x₀).
Thus, the equation of the line passing through the point (0, 0) whose slope’m’ is
(y – 0) = m(x – 0)
⇒ y = mx.

Question 4.
Find the equation of the line passing through (2, 2√3) and inclined with X-axis at an angle of 75°.
Solution:
The slope of the line that inclines with the X-axis at an angle of 75° is m = tan 75°
⇒ m = tan(45° + 30°) = \(\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}\)

= \(\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
We know that the equation of the line passing through the point (x₀, y₀) whose slope m is (y – y₀) = m(x – x₀)
Thus, if a line passes through (2, 2√3) and inclined with the X-axis at an angle of
75°, then the equation of the line is (y – 2√3) = \(=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)(x – 2)
⇒ (y – 2√3)(√3 -1) = (√3 + 1)(x – 2)
⇒ y(√3 – 1) – 2√3(√5 – 1) = x (√3 +1) – 2 (√3 + 1)
⇒ (√3 + 1) x – (√3 -1) y = 2√3 + 2 – 6 + 2√3
⇒ (√3 +1) x – (√3 – 1) y = 4√3 – 4
⇒ (√3 +1) x – (√3 -1) y = 4(√3 – 1)

Question 5.
Find the equation of line intersecting the X-axis at a distance of 3 units to the left of origin with slope -2.
Solution:
It is known that, if a line with slope m makes x-intercept a. Then the equation of the line is y = m(x – a).
For the line intersecting the X-axis at a distance of 3 units to the origin, a = – 3
The slope of the line is given as m = – 2
Thus, the required equation of the line is y = – 2[x – (- 3)]
⇒ y = -2x – 6
⇒ 2x + y + 6 = 0

Question 6.
Find the equation of line intersecting the Y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the X-axis.
Solution:
It is known that, if a line with slope m makes y-intercept c, then, the equation of the line is y = mx + c.
Here, c = 2 and m = tan 30° = \(\frac{1}{2}\)
Thus, the required equation of the line is y = \(\frac{1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ √3y = x + 2√3
∴ x – √3y + 2√3 = 0

Question 7.
Find the equation of the line passing through the points (-1, 1) and (2, -4).
Solution:
It is known that, if the equation of the line passing through the points (x₁; y₁) and (x₂, y₂) is y – y₁ = \(\frac{1}{2}\) (x – x₁)
Therefore, the equation of the line passing through the points (-1, 1) and (2, – 4) is
(y – 1) = \(\frac{-4-1}{2+1}\) (x + 1)
⇒ (y – 1) = \(\frac{-5}{3}\)(x + 1)
⇒ 3(y – 1) = – 5(x + 1)
⇒ 3y – 3 = -5x – 5
∴ 5x + 3y + 2 = 0

Question 8.
The vertices of ΔPQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Solution:
Given that the vertices of ΔPQR are P(2, 1), Q(-2, 3) and R(4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid point of PQ.

By mid point formula, the coordinates of point L is given by \(\left(\frac{2-2}{2}, \frac{1+3}{2}\right)\) = (0, 2)

It is known that, the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\)(x – x₁)

Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2)
Hence, y – 5 = \(\frac{2-5}{0-4}\)(x – 4)
⇒ y – 5 = \(\frac{-3}{-4}\) (x – 4)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x -12
⇒ 3x – 4y + 8 = 0
Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0.

Question 9.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).
Solution:
The slope of the line joining the points (2, 5) and (-3, 6) is m = \(\frac{6-5}{-3-2}=\frac{1}{-5}\)
We know that, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line through the points (2, 5) 1 1
and (- 3, 6) is = \(-\frac{1}{m}=-\frac{1}{\left(-\frac{1}{5}\right)}\) = 5
Now, the equation of the line passing through the point (- 3, 5) whose slope is 5, is
(y – 5) = 5(x + 3)
⇒ (y – 5) = 5x + 15
⇒ 5x – y + 20 = 0

Question 10.
A line perpendicular to the line segment joining the points (1, 0) and (2. 3) divides it in the ratio l:n. Find the equation of the line.
Solution:
According to the section formula, the coordinates of the point that divides the line segment joining the points (1,0) and (2, 3) in the ratio 1 : n is
\(\left(\frac{n(1)+1(2)}{1+

The slope of the line joining the points (1, 0) and (2, 3) is m = [latex]\frac{3-0}{2-1}\) = 3

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other. Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) is \(-\frac{1}{m}=-\frac{1}{3}\).

Now the equation of the line passing through \(\left(\frac{\mathrm{n}+2}{\mathrm{n}+1}, \frac{3}{\mathrm{n}+1}\right)\) and whose slope is \(-\frac{1}{3}\) is (y – \(\frac{3}{(\mathrm{n}+1)}\)) = \(\frac{-1}{3}\)(x – \(\frac{n+2}{n+1}\))
⇒ 3[(n + 1)y – 3] = -[x(n + 1) – (n + 2)]
⇒ 3(n + 1)y – 9 = – (n + 1)x + n + 2
⇒ (1 + n)x + 3(1 + n)y = n + 11

Question 11.
Find the equation of a line that cuts off equal Intercepts on the coordinate axes and passes through the point (2, 3).
Solution:
The equation of a line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1 …….(1)
Here, a and b are the intercepts on X and Y axes respectively. It is given that the line cuts off equal intercepts on both the axes. This means that a = b. Therefore, equation (!) reduces to \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ x + y = a ……….(2)
Since the line passes’through point (2, 3), equation (2) reduces to 2 + 3 = a
a = 5
On substituting the value of ‘a’ in the equation (2), we have x + y = 5
This is the required equation of the line.

Question 12.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight line.
Solution:
Let a,b be the intercepts of the line.
∴ The line cut X – axis at A(a, 0), Y – axis at B (0, b)
Mid point if \(\overline{\mathrm{AB}}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{~b}}{2}\right) \Rightarrow\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{~b}}{2}\right)\) = (2p, 2q)

⇒ \(\frac{a}{2}\) = 2P, \(\frac{b}{2}\) = 2q

⇒ a = 4p, b = 4q
Equation of the line is \(\frac{x}{4 p}+\frac{y}{4 q}\) = 1
⇒ \(\frac{x}{p}+\frac{y}{q}\) = 4

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X – axis measured in the counter clockwise direction.
Solution:
Let ‘θ’ be the inclination.
Slope = -√3
⇒ tan θ = -√3
⇒ θ = \(\frac{2 \pi}{3}\)

Question 14.
The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line, write the value of p in terms of a and b.
Solution:
Equation of the line having intercepts a, b is \(\frac{x}{a}+\frac{y}{b}\) = 1 ⇒ bx + ay – ab = 0 …(1)
p = Length of the perpendicular from (0, 0) to line (1).
= \(\frac{|-a b|}{\sqrt{b^2+a^2}}=\frac{|a b|}{\sqrt{a^2+b^2}}\)

II.

Question 1.
Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
The equation of a line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1 ……. (1)
Here, a and b are the intercepts on X and Y axes respectively, it is given that a + b = 9
⇒ b = 9 – a ……….(2)
From eqns. (1) and (2) we have \(\frac{x}{a}+\frac{y}{9-a}\)
It is given that the line passes through the point (2, 2). Therefore, equation (3) reduces to
\(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ 2\(\left(\frac{1}{a}+\frac{1}{9-a}\right)\) = 1

⇒ \(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ 18 = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3(a – 6) = 0
⇒ (a – 6)(a -3) = 0
⇒ a = 6 or a = 3

If a = 6 then b = 9 – 6 = 3,
Therefore, the equation of the line is \(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ x + 2y – 6 = 0

If a = 3 then b = 9 – 3 = 6, therefore, the equation of the line is
\(\frac{x}{3}+\frac{y}{6}\) = 1
⇒ 2x + y – 6 = 0

Question 2.
Find the equation of the Une through the point (0, 2) making an angle \(\frac{2 \pi}{3}\) with the positive X-axis. Also, find the equation of the line parallel to it and crossing the Y-axis at a distance of 2 units below the origin.
Solution:
The slope of the line making an angle \(\frac{2 \pi}{3}\) with the positive X-axi is m = tan(\(\left(\frac{2 \pi}{3}\right)\)) – √3
Equation of a line passing through (0, 2) and slope √3 is
(y – 2) = -√3(x – 0)
⇒ y – 2 = -√3x
⇒ √3x + y – 2 = 0
The slope of the line parallel to the line √3x + y – 2 = 0 is .4.
It is given that the line parallel to line √3x + y – 2 = 0 crosses the Y – axis 2 units below the origin i.e., it passes through (0, -2).
Hence, the equation of the line passing through the point (0, -2) and having a slope √3 is y – (-2) = -√3(x – 0)
⇒ y + 2 = √3x
⇒ √3x + y + 2 = 0

Question 3.
The perpendicular from the origin to a line meets it at the point (-2, 9). Find the equation of the line.
Solution:
The slope of the line joining the origin (0, 0) and point (-2, 9) is m₁ = \(\frac{9-0}{-2-0}=\frac{-9}{2}\)
Accordingly, the slope of the line perpendicular to the line joining the origin and point (- 2, 9) is m₂ = \(\frac{-1}{m_1}=-\frac{1}{\left(\frac{-9}{2}\right)}=\frac{2}{9}\)
Now, the equation of the line passing through the point (-2, 9) and having a slope ‘2/9’ is
9y – 81 = 2x + 4
⇒ 2x – 9y + 85 = 0

Question 4.
The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Solution:
It is given that when C = 20, the value of L is 124.942
whereas when C = 110, the value of L is 125.134
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, assuming C along the X-axis and L along the Y-axis, we have two points, (20, 124.942) and (110, 125.134) in the xy plane.
Therefore, the linear relation between L and C is the equation of the line passing through the points (20, 124.942) and (110, 125.134)
(L – 124.942) = \(\frac{125.134-124.942}{110-20}\)(C – 20)
L – 124.942 = \(\frac{0.192}{90}\)(C – 20)
L = \(\frac{0.192}{90}\)(C – 20) + 124.942.
This is the required linear relations.

Question 5.
The owner of a milk, store finds that, he can sell 980 litres of milk each week at Rs. 14/ litre and 1220 litres of milk each week at Rs. 16/ litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Solution:
The relationship between selling price and demand is linear. Assuming selling price per litre along the X-axis and demand along the Y-axis, we have two points i.e. (14,980) and (16,1220) in the xy-plane that satisfy the linear relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through the points (14, 980) and (16, 1220).
y – 980 = \(\frac{1220-980}{16-14}\) (x – 14)
⇒ y – 980 = \(\frac{240}{2}\)(x – 14)
⇒ y – 980 = 120(x – 14)
⇒ y = 120(x – 14) + 980
when x = ₹ 17/litre, y = 120(17 – 14) + 980
⇒ y = 120 × 3 + 980 = 360 + 980 = 1340
∴ The owner of the milk store could sell 1340 litres of milk weekly at Rs. 17/ litre.

Question 6.
P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.
Solution:
Let AB be the line segment between the axes and let P(a, b) be its mid-point.
Let the coordinates of A and B are (0, y) and (x, 0) respectively. Since P(a, b) is the mid-point of AB.
\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) = (a, b)
⇒ \(\left(\frac{\mathrm{x}}{2}, \frac{\mathrm{y}}{2}\right)\) = (a, b)
⇒ \(\frac{x}{2}\) = a and \(\frac{y}{2}\) = b
∴ x = 2a and y = 2b

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0). The equation of the line passing through points (0, 2b) and (2a, 0) is
(y – 2b) = \(\frac{(0-2 b)}{(2 a-0)}\)(x – 0)
⇒ y – 2b = \(\frac{1}{2}\)(x)
⇒ a(y – 2b) = -bx
⇒ ay – 2ab = – bx
⇒ bx + ay = 2ab
On dividing both sides by ab, we have \(\frac{b x}{a b}+\frac{a y}{a b}=\frac{2 a b}{a b}\) ⇒ \(\frac{x}{a}+\frac{y}{b}\) = 2
Thus, the equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2

Question 7.
Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.
Solution:
Let AB be the line segment between the axes such that point R(h, k) divides AB in the ratio 1 : 2.
Let the respective coordinates of A and B are (x, 0) and (0, y). Since R(h, k) divides AB in the ratio 1 : 2, according to the section formula
(h, k) = \(\left(\frac{1 \times 0+2 \times x}{1+2}, \frac{1 \times y+2 \times 0}{1+2}\right)\)
⇒ (h, k) = \(\left(\frac{2 \mathrm{x}}{3}, \frac{\mathrm{y}}{3}\right)\)

⇒ h = \(\frac{2 x}{3}\) and k = \(\frac{y}{3}\)

⇒ x = \(\frac{3 h}{2}\) and y = 3k
Therefore, the respective coordinates of A and B are (\(\frac{3 h}{2}\), o) and (0, 3k).
Now, the equation of line AB is
(y – 0) = \(\frac{3 k-0}{0-\frac{3 h}{2}}\) (x – \(\frac{1}{2}\))
⇒ y = \(-\frac{2 \mathrm{k}}{\mathrm{~h}}\)(x – \(\frac{3 \mathrm{h}}{2}\))
⇒ hy = -2kx + 3hk
⇒ 2kx + hy = 3hk
Thus, the required equation of the line is 2kx + hy = 3hk.

Question 8.
By using the concept of equation of a line, prove that the three points are collinear. C
i) (3, 0), (-2, -2) and (8, 2)
Solution:
Let A(3, 0), B(-2, -2), C(8, 2) be the given points.
Equation of the line \(\frac{1}{2}\) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 0 = \(\frac{-2-0}{-2-3}\)(x – 3)
y = \(\frac{2}{5}\)(x – 3)
⇒ 5y = 2x – 6
⇒ 2x – 5y – 6 = 0 …(1)
Put C(8, 2) in (1) ⇒ 2(8) – 5(2) – 6 = 0
Hence given points A, B and C are collinear.

ii) (-5, 1), (5, 5), (10, 7)
Solution:
Let A(-5, 1), B(5, 5), C(10, 7)
Slope of \(\frac{1}{2}\), Slope of \(\frac{1}{2}\)
Slope of \(\frac{1}{2}\) = Slope of \(\frac{1}{2}\)
A, B, C are collinear.
Equation of the line is y – 1 = \(\frac{1}{2}\)(x + 5)
⇒ 5y – 5 = 2x + 10
⇒ 2x – 5y + 15 = 0

iii) (a, b + c), (b, c + a), (c, a + b)
Solution:
Let A(a, b + c), B(b, c + a), C(c, a + b)
Slope of \(\overrightarrow{\mathrm{AB}}=\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}-\mathrm{c}}{\mathrm{~b}-\mathrm{a}}=\frac{\mathrm{a}-\mathrm{b}}{-(\mathrm{a}-\mathrm{b})}\) = -1, Slope of \(\stackrel{\rightharpoonup}{\mathrm{AC}}=\frac{\mathrm{a}+\mathrm{b}-\mathrm{b}-\mathrm{c}}{\mathrm{c}-\mathrm{a}}=\frac{-(\mathrm{c}-\mathrm{a})}{\mathrm{c}-\mathrm{a}}\) = -1

Since slope of \(\overrightarrow{\mathrm{AB}}\) = slope of \(\overrightarrow{\mathrm{AC}}\), the point A, B, C are collinear.
Equation of the line is y – (b + c) = -1(x – a)
⇒ x + y = a + b + c

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9a

Question 1.
Draw a quadrilateral in the cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also find its area.
Solution:
Let ABCD be the given quadrilateral with vertices A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2). Then by plotting A, B, C and D on the cartesian plane and joining AB, BC, CD and DA.The given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly area (ABCD)=Area (ΔABC)+Area (ΔACD)

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by
\(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Therefore, area of
ΔABC = \(\frac{1}{2}\)|- 4(7 + 5) + 0(-5 – 5) + 5(5 – 7)| unit²
= \(\frac{1}{2}\)|- 4(12) + 5(-2)| unit²
= \(\frac{1}{2}\) |- 48 – 10 | unit²
= \(\frac{1}{2}\) × 58 unit²
= 29 unit²

Question 2.
The base of an equilateral triangle with side 2a lies along the Y-axis such that the mid point of the base is at the origin. Find the vertices of the triangle.
Solution:
Let ABC be the given equilateral triangle with side 2a.
Accordingly AB = BC = CA = 2 a
Assume that base BC lies along the Y-axis such that the mid-point of BC is at the origin,
i. e., BO = OC = a, where O is the origin.

Now it is clear that the coordinates of point C are (0, a) while the coordinates of point B are (0, -a). It is known that the line joining a vertex at an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the X-axis.

On applying Pythagoras theorem to ΔAOC, we have (AC)² = (OA)² + (OC)²
(2a)² = (OA)² + a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
Coordinates of point A is (±√3a, 0).
Thus the vertices of the given equilateral triangle are
(0, a)(0, -a) and (√3a, 0) (or) (0, a)(0, -a) and (-√3a, 0)

Question 3.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when
i) PQ is parallel to the Y-axis
ii) PQ is parallel to the X-axis.
Solution:
The given points are P(x₁, y₁) and Q(x₂, y₂)
i) When PQ is parallel to the Y-axis, x₁ = x₂ in this case, distance between P and Q
⇒ \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(y_2-y_1\right)^2}\) = |y₂ – y₁|

ii) When PQ is parallel to the X-axis, y₁ = y₂, in this case, distance between P and Q ⇒ \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-x_1\right)^2}\) = |x₂ – x₁|

Question 4.
Find a point on the X-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let (a, 0) be the point on the X-axis that is equidistant from the points (7, 6) and (3, 4) accordingly
⇒ \(\sqrt{(7-a)^2+(6-0)^2}=\sqrt{(3-a)^2+(4-0)^2}\)

⇒ \(\sqrt{49+a^2-14 a+36}=\sqrt{9+a^2-6 a+16}\)

⇒ \(\sqrt{a^2-14 a+85}=\sqrt{a^2-6 a+25}\)

Squaring on both sides, we have
a² = 14a + 85 = a² – 6a + 25
⇒ -14a + 6a = 25 – 85
⇒ -8a = -60
⇒ a = \(\frac{60}{8}=\frac{15}{2}\)
Thus, the required point on the X-axis is (\(\frac{15}{2}\), 0).

Question 5.
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, -4) and B(8, 0).
Solution:
The coordinates of the mid-point of the line segment joining the points P(0, -4) and B(8, 0) are \(\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)\) = (4, -2)
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is m = \(\frac{y_2-y_1}{x_2-x_1}\), x₂ ≠ x₁
Therefore, the slope of the line passing through (0, 0) and (4, -2) is
= \(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\)
Hence, the required slope of the line is –\(\frac{1}{2}\).

Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
Solution:
Let the given points be A(4, 4), B(3, 5) and C(-1, -1)
Slope of \(\overline{A B}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{3-4}\) = -1

Slope of \(\overline{\mathrm{AC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{-1-4}{-1-4}\) = 1

Slope of \(\overline{\mathrm{AB}}\) × slope of \(\overline{\mathrm{AB}}\) = -1
∴ AB and AC are perpendicular
∴ ΔABC is a right angled triangle

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of Y- axis measured anti — clockwise.
Solution:
If a line makes an angle of 300 with the positive direction of the Y—axis measured anti-clockwise, then the angle made by the line with the positive direction of the
X-axis measured anti-clockwise is 90° + 30° 120°
Thus the slope of the given line is tan 120° = tan(180° – 60°) = – tan 60° = -√3

Question 8.
Without using distance formula, show that points (-2, -1). (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Solution:
Let the points be A (-2, -1), B (4, 0), C (3, 3) and D (-3, 2).
Slope of AB = \(\frac{0+1}{4+2}=\frac{1}{6}\) = and Slope of CD = \(\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}\)
⇒ Slope of AB = Slope of CD ⇒ AB and CD are parallel to each other.
⇒ Slope of BC = Slope of AD ⇒ BC and AD are parallel to each other.
Now, slope of BC = \(\frac{3-0}{3-4}=\frac{3}{-1}\) = -3 and Slope of AD = \(\frac{2+1}{-3+2}=\frac{3}{-1}\) = -3
Therefore, both pairs of opposite sides of a quadrilateral ABCD are parallel. Hence ABCD is a parallelogram.
Thus, points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Question 9.
Find the angle between the X-axis and the line joining the points (3, -1) and (4, -2).
Solution:
The slope of the line joining the points (3, -1) and (4, -2) is m = \(\frac{-2-(-1)}{4-3}\) = -1
Now, the inclination (θ) of the line joining the points (3, -1) and (4, -2) is given by tan θ = -1
⇒ θ = (180° – 45°) = 135°
Thus, the angle between the X-axis and the line joining the points (3, -1) and (4, -2) is 135°.

Question 10.
The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Solution:
Let m₁ and m are the slopes of the two given lines such that m₁ = 2m.
We know that 0 is the angle between the lines l₁ and l₂ with slopes m₁ and m₂ then
tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
It is given that the tangent of the angle between the two lines is 1/3.
\(\frac{1}{3}=\left|\frac{2 m-m}{1+(2 m) m}\right|\)
⇒ \(\frac{1}{3}=\left|\frac{m}{1+2 m^2}\right|\)
⇒ \(\frac{1}{3}=\frac{\mathrm{m}}{1+2 \mathrm{~m}^2}\)
(or)
\(\frac{1}{3}=-\left(\frac{m}{1+2 m^2}\right)=\frac{-m}{1+2 m^2}\)

Case I:
\(\frac{1}{3}=\frac{\mathrm{m}}{1+2 \mathrm{~m}^2}\)
⇒ 2m² + 1 = 3m
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1(m – 1) = 0
⇒ (m – 1)(2m – 1) = 0
⇒ m = 1 or m = -½
If m = 1, then m₁ = 2m = 2(1) = 2
If m = -½, then m₁ = 2(½) = 1

Case II:
\(\frac{1}{3}=\frac{-m}{1+2 m^2}\)
⇒ 2m2 + 2m + m + 1 = 0
⇒ 2m(m + 1) + (m + 1) = 0
⇒ (m + l).(2m + 1) = 0
⇒ m = -1 (or) m = -½
If m = -1, then m₁ = 2m = 2 (-1) = – 2
If m = -½, then m₂ = 2m = 2 (-½) = – 1
Hence the slopes of the lines are 1 and 2 or ½ and 1; -1 and -2 or -½ and -1.

Question 11.
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Solution:
The slope of the line passing through (x₁, y₁) and (h, k) is \(\frac{k-y_1}{h-x_1}\) .
It is given that the slope of the line is m.
∴ \(\frac{k-y_1}{h-x_1}\) = m ⇒ k – y₁ = m(h – x₁).
Hence, k – y₁ = m(h – x₁).

Practice AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Straight Lines MCQ

Question 1.
The equation of the straight line making an intercept of 3 units on the Y -axis and inclined at 45° to the X -axis is
1) y = x – 1
2) y = x + 3
3) 45x + 3
4) y = x + 45
Answer:
2) y = x + 3

Explanation:
θ = 45° (inclined) m = tan 45° = 1;c = 3 ⇒ y = x + 3

Question 2.
The angle made by the line segment joining (5, 2), (6, -15) at (0, 0) is
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) π
Answer:
3) \(\frac{\pi}{2}\)

Explanation:
Let O(0, 0), P(5, 2), Q(6, -15)
m₁ = slope of OP = \(\frac{2}{5}\), m₂ = slope of OQ = \(\frac{-15}{6}=\frac{-5}{2}\), m₁m₂ = -1 ⇒ θ = \(\frac{\pi}{2}\)

Question 3.
The lines 2x + 3y = 6, 2x + 3y = 8 cut the X – axis at A, B respectively. A line T drawn through the point (2, 2) meets the X- axis at ‘C’ in such a way that abscissae A,B,C are in arithmetic progression then the equation of the line T is
1) 2x + 3y = 10
2) 3x + 2y = 10
3) 2x – 3y = 10
4) 3x – 2y = 10
Answer:
1) 2x + 3y = 10

Explanation:
A = (3, 0), B = (4, 0) and let C = (a, 0)
⇒ \(\frac{a+3}{2}\) = 4
⇒ a = 5 (∵ A, B, C are in A.P)
∴ Equation of l is y – 2 = \(\frac{-2}{3}\) (x – 2)

Question 4.
The area (in sq.m) of the triangle formed by the lines x = 0; y = 0 and 3x + 4y = 12 is
1) 3
2) 4
3) 6
4) 12
Answer:
3) 6

Explanation:
\(\frac{C^2}{2|a b|}=\frac{144}{24}\) = 6

Question 5.
The area of the triangle formed by the axes and the line x cos a + y sec a = 2
1) 4
2) 3
3) 2
4) 1
Answer:
3) 2

Explanation:
\(\frac{\mathrm{C}^2}{2|\mathrm{ab}|}=\frac{4}{2|\cos \alpha \cdot \sec \alpha|}\) = 2

Question 6.
If the area of the triangle formed by the lines x =0, y = 0; 3z + 4y = a(a > O) is 1; then a =
1) √6
2) 2√6
3) 4√6
4) 6√2
Answer:
2) 2√6

Explanation:
\(\frac{1}{2}\) = 1
⇒ a = ± \(\sqrt{24}\)
⇒ a = 2√6

Question 7.
EquatIon of the straight line passing through the Image of (3. 6) in the line 2x – y = 5 and perpendlcular to the line 3z +4y = 15 is
1) 4x -3y = 10
2) 4x – 3y = 7
3) 4x – 3y = 5
4) 4x – 3y = 16
Answer:
4) 4x – 3y = 16

Explanation:
Image of (3, 6) is (7, 4)
⇒ Equation of straight line y – y₁ = m(x – x₁) ⇒ 4x – 3y – 16 = 0.

Question 8.
A straight line passing through the point A (3, 4) is such that its Intercept. between the axes bisected at A its equation is
1) 4x+ 3y = 24
2) 3x + 4y = 25
3) x + y = 7
4) 3x – 4y + 7 = 0
Answer:
1) 4x+ 3y = 24

Explanation:
\(\frac{x}{x_1}+\frac{y}{y_1}\) = 2
⇒ \(\frac{x}{3}+\frac{y}{4}\) = 2
⇒ 4x + 3y = 24

Question 9.
If the straight lines y = 4 – 3x; ay = x + 10 and 2y + bx +9 = 10 represet the three consective sides of a rectangle, then ab =
1) 18
2) -3
3) \(\frac{1}{2}\)
4) –\(\frac{1}{3}\)
Answer:
1) 18

Explanation:
m₁ = -3, m₂ = \(\frac{1}{a}\), m₃ = –\(\frac{b}{2}\)
m₁m₂ = -1
⇒ \(\frac{-3}{a}\) = 1
⇒ a = 3;
⇒ m₂ × m₃ = -1
\(\left(\frac{1}{3}\right)\left(\frac{-\mathrm{b}}{2}\right)\) = -1
⇒ b = 6
∴ ab = (3) (6) = 18

Question 10.
The ends of a rod of length ‘1’ moves on the co-ordinate axes the locus of the point on the rod which divides It In the ratIo 1 : 2 is ( 3 )
1) 36x² + 9y² = 4l²
2) 36x² + 9y² = l²
3) 9x² + 36y² = 4l²
4) 9x² + 36y² = l²
Answer:
3) 9x² + 36y² = 4l²

Explanation:
Let A(a, 0), B(0, b) be the end points of the rod.
AB = 1 ⇒ a² + b² = 1² …………… (1)
Let (α, β) divides \(\overrightarrow{\mathrm{AB}}\) in the ratio 1: 2.
∴ (α, β) = \(\left(\frac{2 \mathrm{a}}{3}, \frac{\mathrm{~b}}{3}\right)\)
⇒ a = \(\frac{3 \alpha}{2}\), b = 3β
(1) ⇒ \(\left(\frac{3 \alpha}{2}\right)\)² + (3β)² = l²
∴ Locus of P(a, 13) is 9x² + 36y² = 4l²

Question 11.
Two equal sldes of an isoceles triangle are 7x – y + 3 = 0, x + y – 3 = 0 and its third side passes through the point (1, -10) the equation of the third side is
1) 3x + y + 7 = 0
2) x – 3y + 29 = 0
3) 3x + y + 3 = 0
4) 3x + y – 3 = 0
Answer:
1) 3x + y + 7 = 0

Explanation:
By verification (1, -10) satisfies 3x + y + 7 = 0

Question 12.
The perpendicular distance from (1, 2) to this straight line 12x + 5y = 7 is
1) \(\frac{15}{13}\)
2) \(\frac{12}{13}\)
3) \(\frac{5}{13}\)
4) \(\frac{7}{13}\)
Answer:
1) \(\frac{15}{13}\)

Explanation:
\(\frac{|12+10-7|}{13}=\frac{15}{13}\)

Question 13.
If p and q are the perpendicular distances from the origin to the straight lines x sec θ – y cosec θ = a and x cos θ + y sin θ = a cos 2θ then
1) 4p² + q² = a²
2) p² + q² = a²
3) p² + 2q² = a²
4) 4p² + q² = 2a²
Answer:
1) 4p² + q² = a²

Explanation:
Result only 4p² + q² = a²

Question 14.
The equation of the line parallel to 5x + 12y = 1 and at distance of 3 units from (1, 0) is
1) 5x + 12y + 34 = 0
2) 12x – 5y + 34 = 0
3) 12x – 5y – 44 = 0
4) 5x + 12y + 44 = 0
Answer:
1) 5x + 12y + 34 = 0

Explanation:
By verification (1) and (4) are parallel to the given line
(1) ⇒ \(\frac{|5+0+34|}{13}=\frac{39}{13}\) = 3
⇒ 5x + 12y + 34 = 0 is parallel line which is at a distance of 3 units.

Question 15.
The area (in sq. units) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is
1) 4π
2) 3π
3) 2π
4) π
Answer:
4) π

Explanation:
r = \(\frac{1}{2}\left(\frac{\left|\mathrm{c}_1-\mathrm{c}_2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right)=\frac{1}{2}\left(\frac{|-15+5|}{5}\right)=\frac{1}{2}\)(2) = 1
Area = π(1)² = π

Question 16.
If A (2, -1) and B (6, 5) are two points, the ratio in which the foot of the perpendicular from (4, 1) to AB divides It in
1) 8 : 15
2) 5 : 8
3) -5 : 8
4) -8 : 5
Answer:
2) 5 : 8

Explanation:
Slope of AB = and Slope of the line ⊥^r to \(\overline{\mathrm{AB}}=-\frac{2}{3}\)
Equation is y – 1 = (x – 4) ⇒ 2x + 3y – 11 = 0
Ratio = \(\frac{-(4-3-11)}{(12+15-11)}=\frac{10}{16}=\frac{5}{8}\) = 5 : 8

Question 17.
A straight line which makes equal intercepts on positive X and Y axes and which is at a distance 1 unit from the origin intersects the straight line
y = 2x + 3 + √2 at (x₀, y₀) then 2x₀ + y₀ =
1) 3 + √2
2) √2 – 1
3) 1
4) 0
Answer:
2) √2 – 1

Explanation:
Let 1 be the line y- a.
Perpendicular distance from origin is 1 ⇒ \(\frac{|\mathrm{a}|}{\sqrt{2}}\) = 1 a = √2
Now equation is x + y = √2 …………. (1) and Given line y = 2x + 3 + √2 …………. (2)
From (1) and (2) we get x = -1 and y = √2 + 1;
x₀ = -1, y₀ = √2 + 1 2x₀ + y₀ = -2 + √2 = √2 – 1

Question 18.
If a, b, c are in AP, the line ax + by + c = 0 passes through the fixed point
1) (1, 2)
2) (-1, 2)
3) (1, -2)
4) (-1, -2)
Answer:
3) (1, -2)

Explanation:
If a, b, c are in AP then \(\frac{\mathrm{a}+\mathrm{c}}{2}\) = b
a – 2b + c = 0 →, ax + by + c = 0; (x, y) = (1, -2)

Question 19.
If a, b, c are in AP then the line ax + by + c = 0
1) passes through a fixed point
2) form an equilateral triangle
3) form a rhombus
4) form a square
Answer:
1) passes through a fixed point

Explanation:
From (18) the line passing through fixed point.

Question 20.
The point on the line 2x – 3y = 5 which is equidistant from (1, 2) and (3, 4) is
1) (2, 3)
2) (4, 1)
3) (1, -1)
4) (4, 6)
Answer:
2) (4, 1)

Explanation:
By verification option (2) and (3) satisfies the 2x – 3y = 5 and distance from (4, 1) is
\(\sqrt{9+1}=\sqrt{10}\) and \(\sqrt{1+9}=\sqrt{10}\)
Option (2) satisfies.

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8b

I.

Question 1.
Find the 20^th and n^th terms of the G.P. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}\) ………………
Solution:
Given series is \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}\) ………….
Here, the first term a = \(\frac{5}{2}\) and the common ratio r = \(\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{2}{4}=\frac{1}{2}\)

a_n = ar^n-1
⇒ a₂₀ = ar¹⁹
= \(\left(\frac{5}{2}\right)\left(\frac{1}{2}\right)^{19}=\frac{5}{2^{20}}\) and

a_n = ar^n-1
= \(\left(\frac{5}{2}\right)\left(\frac{1}{2}\right)^{\mathrm{n}-1}=\frac{5}{2^{\mathrm{n}}}\)

Question 2.
Find the 12^th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
Common ratio, r = 2, let a be the first term of the G.P.
∴ a₈ = ar^8-1 = ar⁷
⇒ ar⁷ = 192
⇒ a(2)⁷ = (2)⁶(3)
⇒ a = \(\frac{2^6 \times 3}{2^7}=\frac{3}{2}\)
Therefore, a₁₂ = ar¹¹
2¹¹ = 3 × 2¹⁰ = 3072

Question 3.
The 5^th, 8^th and 11^th terms of a G.P are p,q and s, respectively. show that q² = ps.
Solution:
Let a be the first term and r be the common ratio of the G.P
According to the given condition,
a₅ = a^5-1 = ar⁴ = p …………..(1)
a₈ = a^8-1 = ar⁷ = p …………..(2)
a₁₁ = a^11-1 = ar¹⁰ = p …………..(3)
Now, multiplying Eqs. (1) and (3)
ar⁴ . ar¹⁰ = ps
⇒ a²r¹⁴ = ps
⇒ (ar⁷)² = ps
⇒ q² = ps [∵ from (2)]

Question 4.
The 4th term of a G.P. is square of its second term, and the first term is -3. Determine its 7^th term.
Solution:
Let a be the first term and r be the common ratio of the G.P. ∴ a = -3
Given a₄ = (a₂)² ⇒ ar³ = (ar)² ⇒ ar³ = a²r² ⇒ r = a ⇒ r = -3
Therefore, a₇ = ar^7-1 = ar⁶ = (-3)(-3)⁶ = -(3)⁷ = -2187
Hence, the seventh term of the G.P. is -2187.

Question 5.
Which term of the following sequences,
a) 2, 2 √2 ,4…is 128 ?
Solution:
Given that 2, 2√2 ,4…
Here, a = 2 and r = \(\frac{1}{2}\) = √2
Let a_n = 128
⇒ ar^n-1 = 128
⇒ 2(√2)^n-1 = 2⁶
⇒ 2^{\(\frac{1}{2}\)} = 2⁶
⇒ \(\frac{n-1}{2}\) = 6
⇒ n = 13
Hence, the 13^th term of the given sequence is 128.

b) √3, 3, 3√3 …. is 729 ?
Solution:
√3, 3, 3√3 ….
Given that √3, 3, 3√3 …. 729
Here, a = √3 & r = \(\frac{3}{\sqrt{3}}\) = √3
Let a_n = 729
⇒ ar^n-1 = 729
⇒ √3(√3)^n-1 = 3⁶
⇒ (√3)ⁿ = 3⁶
⇒ (3)^{\(\frac{n}{2}\)} = 3⁶
⇒ \(\frac{n}{2}\) = 6
⇒ n = 12
Hence, the 12^th term of the given sequence is 729.

c) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\), ……… is \(\frac{1}{19683}\)?
Solution:
\(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\), ………
Given \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\), ………

Here, a = \(\frac{1}{3}\) and r = \(\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{3}{9}=\frac{1}{3}\)

Let a_n = \(\frac{1}{19683}\)

⇒ ar^n-1 = \(\frac{1}{19683}\)

⇒ \(\frac{1}{3}\left(\frac{1}{3}\right)^{\mathrm{n}-1}=\left(\frac{1}{3}\right)^9\)

⇒ \(\left(\frac{1}{3}\right)^{\mathrm{n}}=\left(\frac{1}{3}\right)^9\)

⇒ n = 9
Hence, the 9^th term of the given sequence is \(\frac{1}{19683}\)

Question 6.
For what values of x, the numbers –\(\frac{2}{7}\), x, –\(\frac{7}{2}\) are in G.P. ?
Solution:
Here, –\(\frac{2}{7}\), x, –\(\frac{7}{2}\) are in G.P.
⇒ \(\frac{x}{\frac{2}{-7}}=\frac{\frac{-7}{2}}{x}\)
⇒ x² = \(\frac{2}{7} \times \frac{7}{2}\) = 1
⇒ x = ± 1
Hence, for x = ± 1, the given numbers will be in GP.

II. Find the sum to indicated number of terms in each of the G.P in Exercises 1 to 4 :

Question 1.
0.15, 0.015, 0.0015, … 20 terms.
Solution:
Here, a = 0.15 and r = \(\frac{0.015}{0.15}\) = 0.1;
S_n = \(\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}\)
⇒ 0.1 < 1 and n = 20
⇒ S₂₀ = \(\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}=\frac{0.15\left[1-(0.1)^{20}\right]}{0.9}=\frac{1}{6}\)[1 – (0.1)²⁰]

Question 2.
√7, √21, 3√7 , …. n terms.
Solution:
Here, a = √7 and r = \(\frac{\sqrt{21}}{\sqrt{7}}\) = √3; S_n = \(\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}\) [∵ r > 1]
⇒ S_n = \(\frac{\sqrt{7}\left[(\sqrt{3})^{\mathrm{n}}-1\right]}{\sqrt{3}-1}=\frac{\sqrt{7}\left[(\sqrt{3})^{\mathrm{n}}-1\right]}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

\(\frac{\sqrt{7}(\sqrt{3}+1)\left[3^{\frac{\mathrm{n}}{2}}-1\right]}{2}=\frac{\sqrt{7}}{2}\)(√3 + 1)[3^{\(\frac{\mathrm{n}}{2}\)} – 1]

Question 3.
1, -a, a², -a³, ….. n terms (if a ≠ -1).
Solution:
Here, a = 1 and r = \(\frac{1}{2}\) = -a;
⇒ S_n = a\(\left(\frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}\right)\)

⇒ S_n = \(\frac{\left[1-(-a)^n\right]}{1-(-a)}=\frac{1-(-a)^n}{1+a}\)

Question 4.
x³, x⁵, x⁷, …………. n terms (if x ≠ 1).
Solution:
Here, a = x³ and r = \(\frac{x^5}{x^3}\) = x²;
⇒ S_n = \(\frac{\mathrm{a}\left[1-\mathrm{r}^{\mathrm{n}}\right]}{1-\mathrm{r}}\)
S_n = \(\frac{x^3\left[1-\left(x^2\right)^n\right]}{1-x^2}=\frac{x^3\left[1-x^{2 n}\right]}{1-x^2}\)

Question 5.
Evaluate \(\sum_{k=1}^{11}\)(2 + 3^k)
Solution:
\(\sum_{k=1}^{11}\)(2 + 3^k) = (2 + 3¹) + (2 + 3²) + (2 + 3³) + ………….. + (2 + 3¹¹)
= 2 × 11 + (3¹ + 3² + 3³ + ………… + 3¹¹)
= 22 + \(\frac{3\left(3^{11}-1\right)}{3-1}\)
= 22 + \(\frac{3}{2}\)(3¹¹ – 1)

Question 6.
The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.
Solution:
Let a, ar and ar² be three terms of a GP.
According to the question, sum of first three terms a + ar + ar² = \(\frac{39}{10}\) …………. (1)
Product of terms a × ar × ar² = 1
⇒ a³r³ = 1
⇒ ar = 1 ………. (2)
Dividing equation (1) by (2), we have
\(\frac{a+a r+a r^2}{a r}=\frac{39}{10}\)

⇒ \(\frac{1+r+r^2}{r}=\frac{39}{10}\)

⇒ 10 + 10r + 10r² = 39r
⇒ 10r² – 29r + 10 = 0
⇒ 10r² – 25r – 4r + 10 = 0
⇒ 5r(2r – 5) – 2(2r – 5) = 0
⇒ (5r – 2) (2r – 5) = 0
⇒ r = \(\frac{2}{5}\) (or) \(\frac{5}{2}\)

If r = \(\frac{2}{5}\), from (2), we have a × \(\frac{2}{5}\) = 1
⇒ a = \(\frac{5}{2}\)

Therefore, the three terms are \(\frac{5}{2}\), 1, \(\frac{2}{5}\)

If r = \(\frac{5}{2}\) from (2) we have a = \(\frac{2}{5}\).
The three terms are \(\frac{2}{5}\), 1, \(\frac{5}{2}\)

Question 7.
How many terms of G.P. 3, 3², 3³, …. are needed to give the sum 120 ?
Solution:
The given G.P. is 3, 3², 3³, ….
Here a = 3 and r = \(\frac{3^2}{3}\) = 3 3
Let n terms of this G.P. be required to obtain the sum as 120.
⇒ S_n = \(\frac{3\left(3^n-1\right)}{3-1}\) = 120
⇒ 3(3ⁿ – 1) = 240
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81
⇒ 3ⁿ = 3⁴
⇒ n = 4
Hence, four terms of the given G.P. are required to obtain the sum as 120.

Question 8.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum of n terms of the GP.
Solution:
Let the G.P. be a, ar, ar², ar³ ………….
Given a + ar + ar² = 16
⇒ a(1 + r + r²) = 16 ……….. (1)
ar³ + ar⁴ + ar⁵ = 128
⇒ ar³(1 + r + r²) = 128 …………. (2)

Dividing equation (2) by (1), we obtain \(\frac{1}{2}\)
⇒ r³ = 8
⇒ r = 2

Putting the value of r in equation (1), we have
a(1 + 2 + 2²) = 16
⇒ 7a = 16
⇒ a = \(\frac{16}{7}\)

Now, S_n = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)

⇒ S_n = \(\frac{\frac{16}{7}\left(2^n-1\right)}{2-1}=\frac{16}{7}\)(2ⁿ – 1)

Question 9.
Given a G.P, with a = 729 and 7^th term 64, Determine S₇
Solution:
Given that a = 729 and 7^th term 64
⇒ a₇ = ar^7-1 = 64
⇒ r⁶ = \(\frac{64}{729}\)
⇒ r⁶ = \(\left(\frac{2}{3}\right)^6\)
⇒ r = \(\frac{2}{3}\) < 1

now S_n = \(\frac{a\left(1-r^n\right)}{1-r}\)

⇒ S_n = \(\frac{729\left[1-\left(\frac{2}{3}\right)^7\right]}{1-\frac{2}{3}}=\frac{729\left[1-\left(\frac{2}{3}\right)^7\right]}{\frac{1}{3}}\)

= 729 × 3[1 – \(\frac{128}{2187}\)]

= 2187\(\left[\frac{2187-128}{2187}\right]\)
= 2187 – 128
= 2059

Question 10.
Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.
Solution:
Let the first term of a given G.P is a and the common ratio is r.
Sum of the first two terms is -4, therefore, a₁ + a₂ = -4
⇒ a + ar = -4 ……… (1)
The fifth term is 4 times to the third term, therefore,
a₅ = 4a₃
⇒ ar⁴ = 4 x ar
⇒ r² = 4
⇒ r = ± 2
If r = 2, from (1), we have a + a × 2 = -4
⇒ 3a = -4
⇒ a = \(-\frac{4}{3}\)
Therefore, the required GP : a, ar, ar², …. is given by \(-\frac{4}{3},-\frac{4}{3} \times 2,-\frac{4}{3} \times 2^2\) ….. or \(-\frac{4}{3},-\frac{8}{3},-\frac{16}{3}\) …………..
If r = -2, from (1), we have a + a × (-2) = -4 ⇒ -a = -4 ⇒ a = 4
Therefore, the required GP : a ar ar², ……… is given by 4, 4 × (-2), 4 × (-2)², …….. or 4, -8, 16 …………

Question 11.
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Solution:
According to the given condition,
a₄ = ar³ = x ……….. (1)
a₁₀ = ar⁹ = y ………..(2)
a₁₆ = ar¹⁵ = z …… (3)
Now multiplying eq (1) by (3) ⇒ ar³ . ar¹⁵ = x × z
⇒ a²r¹⁸ = xz
⇒ (ar⁹)² = xz ⇒ y² = xz [∵ from (2)]
∴ x, y, z are .in G.P

Question 12.
Find the sum of n terms of the sequence, 8, 88, 888, 8888 ….
Solution:
The given sequence is 8, 88, 888, 8888 …………..
This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as
S_n = 8 + 88 + 888 +8888 + ………… to n terms
⇒ \(\frac{8}{9}\) [9 + 99 + 999 + 9999 + …. to n terms]
⇒ \(\frac{8}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + (10000 – 1) + ………… to n terms]
⇒ \(\frac{8}{9}\) [(10 + 10² + 10³ + 10⁴ + …… + 10ⁿ) – (1 + 1 + 1 + 1 + ……….. to n terms]
⇒ \(\frac{8}{9}\) [(10¹ + 10² + 10³ + 10⁴ + ………. + 10ⁿ) – n]

⇒ \(\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]=\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{9}\right]-\frac{8}{9}\)n = \(\frac{80}{81}\)(10ⁿ – 1) – \(\frac{8}{9}\)n

Question 13.
Find the sum of the products of the corresponding terms of the two sequences
2. 4, 8, 16, 32 and 128, 32, 8, 2, ½.
Solution:
Given sequences are 2, 4, 8, 16, 32 … (1)
128, 32, 8, 2, \(\frac{1}{2}\) … (2)
Multiplying the eq (1) and (2), we get 256, 128, 64, 32, 16
Let Sn = 256 + 128 + 64 +32 + 16,
Here a = 256, r = \(\frac{1}{2}\) < 1,
Now, S_n = \(\frac{a\left(1-r^n\right)}{1-r}\)
S_n = \(\frac{256\left[1-\left(\frac{1}{2}\right)^5\right]}{1-\frac{1}{2}}=\frac{256\left[1-\frac{1}{32}\right]}{\frac{1}{2}}\)
= 512\(\left[\frac{32-1}{32}\right]\)
= 16 × 31
= 496

Question 14.
Show that the products of the corresponding terms of the sequences a, ar, ar² ………… ar^n-1 and A, AR, AR², ‚……………, AR^n-1 form a GP. and find the common ratio.
Solution:
Given sequences are a, ar, ar², ……. ar^n-1 …………….. (1)
and A, AR¹, AR² … AR^n-1 ……….. (2)
Now multiplying the corresponding terms of eq (1) and (2), we get
aA, arAR¹, ar²AR² ……. ar^n-1 . AR^n-1
Common ratio = \(\frac{\text { Second term }}{\text { First term }}=\frac{\mathrm{aArR}}{\mathrm{aA}}\) = rR

III.

Question 1.
Find tour numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.
Solution:
Let a be the first term and r be the common ratio of the G.P.
a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³
Bythe given condition, a₃ = a₁ + 9 ar² = a + 9 …………. (1)
and a₂ = a₄ + 18 = ar = ar³ + 18 ………. (2)
From (1) & (2), we have a(r² – 1) = 9 …………… (3)
ar(1 – r²) = 18 ……….. (4)
Dividing (4) by (3), we have \(\frac{a r\left(1-r^2\right)}{a\left(r^2-1\right)}=\frac{18}{9}\)
⇒ -r = 2
⇒ r = -2
Substituting the value of r in (1), we obtain 4a = a + 9
⇒ 3a = 9
⇒ a = 3
Hence, the first four numbers of the GP are 3, 3(-2).3(-2)² and 3(-2) i.e. 3, -6, 12 & -24.

Question 2.
If the pth, qth and rth terms of a G.P. are a, b and c respectively. Prove that a^q-r b^r-p c^p-q = 1.
Solution:
Let A be the first term and R be the common ratio of the GP.
According to the question, a_p = AR^p-1 = a …………(1)
a_q = AR^q-1 = b ………(2)
a_r = AR^r-1 = c ………..(3)
Now, LHS = a^q-rb^r-pc^p-q = (AR^p-1)^q-r(AR^q-1)^r-p(AR^r-1)^p-q
= A^q-rR^{(p – 1)(q – r)}A^{r – p}R^(q-1)(r-p)A^p-qR^{(r – 1)(P – 1)}
= A^{q – r + r – p + p – q} × R^{(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)}
= A⁰ × R^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}
= A°R° = 1 = RHS

Question 3.
If the first and the nth term of a G.P. are a and b respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.
Solution:
Here, a is the first term, let the common ratio be r.
According to the problem, b = ar^n-1
Now, P = a × ar × ar² × ………. × ar^n-1 = aⁿ × (r.r².r³…r^n-1)
= aⁿr^{1+2+3+….+(n-1)}
= aⁿr^{\(\frac{\mathrm{n}-1}{2}\)[2 × 1 + [(n-1)-1)1]}
⇒ P = aⁿr^{\(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)}

Therefore, P² = [aⁿr^{\(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)}]² = a²ⁿ. r^n(n-1)
= aⁿ. aⁿ. r^n(n-1)
= aⁿ . (a . r^n-1)ⁿ
= aⁿbⁿ = (ab)ⁿ

Question 4.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is
Solution:
Let a be the first term and the common ratio be r. Now, the ratio of the sum of first
n terms of a GP to the sum of terms from (n+ 1)^th to (2)^th term is given by
\(\frac{S_n}{S_{2 n}-S_n}=\frac{\frac{a\left(r^{11}-1\right)}{r-1}}{\frac{a\left(r^{2 n}-1\right)}{r-1}-\frac{a\left(r^n-1\right)}{r-1}}\)

Question 5.
If a,b, c and d are in G.P. showthat(a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let the common ratio be r. According to the question, b = ar, c = ar² and d = ar³
LHS = (a² + b² + c²) (b² + c² + d²) = [a² + (ar)² + (ar²)²][(ar)² + (ar²)² + (ar³)²]
= [a² + a²r² + a²r⁴] [a²r² + a²r⁴ + a²r⁶] — [a²[1 + r² + r⁴] = a²r²[1 + r² + r⁴] = a⁴r²[1 + r² + r⁴]² ……………. (1)
RHS = (ab + bc + cd)² = [a(ar) + (ar)(ar²) + (ar²)(ar³)]² = [ar + a²r³ + a²r⁵]²
= [a²r(1 + r² + r⁴)]² = a⁴r²[1 + r² + r⁴]² ………. (2)
From (1) and (2), we have (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

Question 6.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G₁ & G₂ be two numbers between 3 & 81 such that the series 3, G₁, G₂, 81 forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴ 81 = 3.r³ = r³ = 27
∴ r = 3 (Taking real roots only)
If r = 3, G₁ = ar = (3)(3) = 9 & G₂ = ar² = (3)(3)² = 27.
Thus, the required two numbers are 9 & 27.

Question 7.
Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\) may be the geometric mean between a and b.
Solution:
According to the question, \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{a b}\)
⇒ \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\) = a^½b^½
⇒ aⁿ⁺¹ + bⁿ⁺¹ = a^½b^½(aⁿ + bⁿ)
⇒ aⁿ⁺¹ + bⁿ⁺¹ = a^n+½b^½ + a^½b^n+½ = 0
⇒ aⁿ⁺¹ – a^n+½b^½ + bⁿ⁺¹ – a^½b^n+½ = 0
⇒ a^n+½(a^½ – b^½) – b^n+½(a^½ – b^½) = 0
⇒ (a^½ – b^½)(a^n+½ – b^n+½) = 0
⇒ (a^n+½ – b^n+½) = 0
⇒ a^n+½ = b^n+½
⇒ \(\frac{a^{n+1 / 2}}{b^{n+1 / 2}}\) = 1
⇒ \(\left(\frac{a}{b}\right)^{n+1 / 2}\) = 1
⇒ \(\left(\frac{a}{b}\right)^0\)
∴ n + \(\frac{1}{2}\) = 0
⇒ n = –\(\frac{1}{2}\)

Question 8.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2): (3 – 2√2).
Solution:
Let a and b are the two numbers whose geometric mean is √ab .
According to the question, a + b = 6√ab ……..(1)
⇒ (a + b)² = 36ab
⇒ a² + b² + 2ab = 36ab
⇒ a² + b² = 34ab
⇒ a² + b² – 2ab = 32ab
⇒ (a – b)² = 32ab
⇒ a – b = 4 √2√ab ………….. (2)
Adding equations (1) & (2) we have,
2a = 6√ab + 4√2 √ab

⇒ a = (3 + 2√2)√ab ……….. (3)
Subtracting equations (1) and
2b = 6√ab – 4√2√ab

⇒ b = (3 – 2√2)√ab ………(4)
From the eq (3) and (4),
∴ \(\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\)

Question 9.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± \(\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\)
Solution:
It is given that A & G are AM & GM between two positive numbers.
Let these two positive numbers be a & b.
AM = A = \(\frac{1}{2}\) ⇒ a + b = 2A ……… (1)
GM = G = √ab ⇒ ab = G² ……… (2)
Substituting the values of (a + b) and (ab) from (1) & (2) in the identity
(a – b)² = (a + b)² – 4ab, we have (a – b)² = 4A² – 4G² = 4(A² – G²) = 4(A – G)(A + G)
⇒ a – b = 2\(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) ……….. (3)
Adding (1) & (3) we have 2a = 2A + 2\(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\)
⇒ a = A + \(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\)
Substituting the value of a in (1) we have
A + \(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) + b = 2A
⇒ b = A – \(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\)
Hence, the two numbers a & b are A ± \(\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\)

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present In the culture originally, how many bacteria will be present at the end of 2 hour, 4th hour and nth hour?
Solution:
It is given that the no. of bacteria doubles every hour therefore, the no. of bacteria after every hour will form a G.P.
Here, a = 30 & r = 2
∴ a³ = ar² = (30)(2)² = 120,
a₅ = ar⁴ = (30) (2)⁴ = 30 × 16 = 480
Therefore, the no. of bacteria at the end of 2nd hour will be 120.
Therefore, the no. of bacteria at the end of 4th hour will be 480.
aⁿ⁺¹ = arⁿ = (30) 2ⁿ.
Hence, no. of bacteria at the end of n^th hour will be (30)2ⁿ.

Question 11.
What will Rs. 500 amounts in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually ?
Solution:
The amount deposited in the bank is Rs. 500.
At the end of first year, amount
At the end of 2nd year, amount = ₹ 500 (1.1) (1.1)
At the end of 3rd year, amount = ₹ 500 (1.1) (1.1) (1.1) and so on
∴ Amount at the end of 10 years = ₹ 500 (1.1) (1.1) …. (10 times)

Question 12.
If AM and GM of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution:
Let the roots of the quadratic equation be a & b.
According to the question, AM = \(\frac{1}{2}\) = 8
⇒ a + b = 16 …….(1)
GM = \(\frac{1}{2}\) = 5 ⇒ ab = 25 ………….(2)
The quadratic equation is given by
x² – x (sum of roots) + (product of roots) = 0
⇒ x² – x(a+b) + (ab) = 0
⇒ x² – 16x + 25 = 0 [Using (1) 8s (2)]
Therefore, the required quadratic equation is x² – 16x + 25 = 0.

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8a

I. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are given.

Question 1.
a_n = n(n + 2)
Solution:
Given that a_n = n(n + 2).
Substituting n = 1, 2, 3, 4 and 5, we have
a₁ = 1(1 + 2) = 3; a₂ = 2(2 + 2) = 8; a₃ = 3(3 + 2) = 15; a₄ = 4(4 + 2) = 24; a₅ = 5(5 + 2) = 35
Hence, the required terms are 3, 8, 15, 24 and 35.

Question 2.
a_n = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
Solution:
Given that a_n = \(\frac{\mathrm{n}}{\mathrm{n}+1}\)
Substituting n = 1, 2, 3, 4 and 5 we have,
a₁ = \(\frac{1}{1+1}=\frac{1}{2}\); a₂ = \(\frac{2}{2+1}=\frac{2}{3}\); a₃ = \(\frac{3}{3+1}=\frac{3}{4}\); a₄ = \(\frac{4}{4+1}=\frac{4}{5}\); a₅ = \(\frac{5}{5+1}=\frac{5}{6}\)
Hence, the required terms are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}\) & \(\frac{5}{6}\).

Question 3.
a_n = 2ⁿ
Solution:
Given that a_n = 2ⁿ
Substituting n = 1, 2, 3, 4 and 5 we have
a₁ = 2¹ = 2, a₂ = 2² = 4, a₃ = 2³ = 8, a₄ = 2⁴ = 16, a₅ = 2⁵ = 32
Hence, the required terms are 2, 4, 8, 16 and 32.

Question 4.
a_n = \(\frac{2 n-3}{6}\)
Solution:
Given that a_n = \(\frac{2 n-3}{6}\)
Substituting n = 1, 2, 3, 4 and 5 we have
a₁ =(-1)^1-1.5¹⁺¹ = (-1)⁰.5² = 25,
a₂ =(-1)^2-1.5²⁺¹ = (-1)¹.5³ = 25,
a₃ =(-1)^3-1.5³⁺¹ = (-1)².5⁴ = 25,
a₄ =(-1)^4-1.5⁴⁺¹ = (-1)³.5⁵ = 25,
a₅ =(-1)^5-1.5⁵⁺¹ = (-1)⁴.5⁶ = 25,
Hence, the required terms are 25, -125, 625, -3125 & 15625

Question 6.
a_n = n\(\frac{n^2+5}{4}\)
Solution:
Given that a_n = n\(\frac{n^2+5}{4}\)
Substituting n = 1, 2, 3, 4 and 5 we have
a₁ = 1 × \(\left[\frac{1^2+5}{4}\right]\) = 1 × \(\left[\frac{2^2+5}{4}\right]\)

a₂ = 2 × \(\left[\frac{3^2+5}{4}\right]\) = 1 × \(\frac{9}{4}=\frac{9}{2}\)

a₃ = 3 × \(\left[\frac{4^2+5}{4}\right]\) = 1 × \(\frac{14}{4}=\frac{21}{2}\)

a₄ = 4 × \(\left[\frac{5^2+5}{4}\right]\) = 1 × \(\frac{21}{4}\) = 21

a₅ = 5 × \(\left[\frac{5^2+5}{4}\right]\) = 1 × \(\frac{30}{4}=\frac{75}{2}\)
Hence, the required terms are \(\frac{3}{2}, \frac{9}{2}, \frac{21}{2}\), 21 and \(\frac{75}{2}\)

Find the indicated terms in each of the sequences In Exercises 7 to 10 whose n^th terms are given.

Question 7.
a_n = 4n – 3;(a₁₇, a₂₄)
Solution:
Given that a_n = 4n – 3
Substituting n = 17, we have a₁₇ = 4(17) – 3 = 68 – 3 = 65
Substituting n = 24, we have a₂₄ = 4(24) – 3 = 96 – 3 = 93

Question 8.
a_n = \(\frac{\mathrm{n}^2}{2^{\mathrm{n}}}\), a₇.
Solution:
Given that a_n = \(\frac{\mathrm{n}^2}{2^{\mathrm{n}}}\).
Substituting n = 7,we have a₇ = \(\frac{7^2}{2^7}=\frac{49}{128}\)

Question 9.
a_n = (-1)^n-1n³, a₉
Solution:
Given that a_n = \(\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3}\)
Substituting n = 9, we have, a₉ = (-1)^9-1. 9³ = (-1)⁸ . 9³ = 9³ = 729

Question 10.
a_n = \(\frac{n(n-2)}{n+3}\);a₂₀
Solution:
Given that a = \(\frac{n(n-2)}{n+3}\)
Substituting n = 20, we have a₂₀ = \(\frac{20(20-2)}{20+3}=\frac{20 \times 18}{23}=\frac{360}{23}\)

II. Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series.

Question 1.
a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1
Solution:
Given that a₁ = 3, a_n = 3a_n-1 + 2 for all n> 1
Substituting n = 2, 3, 4 and 5, we have
a₂ = 3a_2-1 + 2 = 3a₁ + 2 = 3 × 3 + 2 = 11,
a₃ = 3a_3-1 + 2 = 3a₂ + 2 = 3 × 11 + 2 = 11,
a₄ = 3a_4-1 + 2 = 3a₃ + 2 = 3 × 35 + 2 = 11,
a₅ = 3a_5-1 + 2 = 3a₄ + 2 = 3 × 107 + 2 = 11,
Therefore, the required terms are 3, 11, 35, 107 and 323.
Hence, the corresponding series is 3 + 11 +35 + 107 + 323 + ………..

Question 2.
a₁ = -1, a_n = \(\frac{a_{n-1}}{n}\); n ≥ 2
Solution:
Given that a₁ = -1, a_n = \(\frac{\mathrm{a}_{\mathrm{n}-1}}{\mathrm{n}}\), n ≥ 2

Substituting n = 2, 3, 4 and 5 we have,
a₂ = \(\frac{a_{2-1}}{2}=\frac{a_1}{2}=-\frac{1}{2}\)

a₃ = \(\frac{a_{3-1}}{3}=\frac{a_2}{3}=\frac{-\frac{1}{2}}{3}=\frac{-1}{6}\)

a₄ = \(\frac{a_{4-1}}{4}=\frac{a_3}{4}=\frac{-\frac{1}{6}}{4}=\frac{-1}{24}\)

a₅ = \(\frac{a_{5-1}}{5}=\frac{a_4}{5}=\frac{-\frac{1}{24}}{5}=\frac{-1}{120}\)

Hence, the required terms are -1, \(-\frac{1}{2},-\frac{1}{6},-\frac{1}{24}\) & \(-\frac{1}{120}\)

Hence, the corresponding series is -1 + \(\left(-\frac{1}{2}\right)+\left(-\frac{1}{6}\right)+\left(-\frac{1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots \ldots\)

Question 3.
a₁ = a₂ = 2; a_n = a_n-1 – 1, n > 2
Solution:
Given that a₁ = a₂ = 2; a_n = a_n-1 – 1, n > 2
Substituting n = 3, 4 and 5 we have
a₃ = a_3-1 – 1 = a₂ – 1 = 2 – 1 = 1
a₄ = a_4-1 – 1 = a₃ – 1 = 1 – 1 = 1
a₅ = a_5-1 – 1 = a₄ – 1 = 0 – 1 = 1
Hence, the required terms are 2, 2, 1,0 and -1
Hence, the corresponding series is 2 + 2+ 1 + 0 + (-1) ….

Question 4.
The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_n-1 + a_n-2, n > 2.
Find \(\frac{a_{n+1}}{a_n}\), for n = 1, 2, 3, 4, 5.
Solution:
Given that 1 = a₁ = a₂ and a_n = a_n-1 + a_n-2

Substituting n = 3, 4 and 5, we have
a₃ = a_3-1 + a_3-2 = a₂ + a₁ = 1 + 1 = 2
a₄ = a_4-1 + a_4-2 = a₃ + a₂ = 2 + 1 = 2
a₅ = a_5-1 + a_5-2 = a₄ + a₃ = 3 + 2 = 2
a₆ = a_6-1 + a_6-2 = a₅ + a₄ = 5 + 3 = 2

Putting n = 1 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}\) = 1

Putting n = 2 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}\) = 2

Putting n = 3 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}\)

Putting n = 4 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}\)

Putting n = 5 in \(\frac{a_{n+1}}{a_n}\), we have \(\frac{a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8d

Find the sum to infinity in each of the following Geometric Progression.

Question 1.
1, \(\frac{1}{3}, \frac{1}{9}\), …………..
Solution:
Let S = 1, \(\frac{1}{3}, \frac{1}{9}\), …………..; Here, a = 1 and r = \(\frac{1}{3}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{1}{1-\frac{1}{3}}=\frac{1}{2 / 3}=\frac{3}{2}\) = 1.5

Question 2.
6, 1.2, 0.24, …………………
Solution:
Let S = 6 + 1.2 + 0.24 + ………………….; Here, a = 6 and r = 0.2
So, S_∞ = \(\frac{a}{1-r}=\frac{6}{1-(0.2)}=\frac{6}{0.8}\) = 7.5

Question 3.
5, \(\frac{20}{7}, \frac{80}{49}\) ………..
Solution:
Let S = 5, \(\frac{20}{7}, \frac{80}{49}\), ………….; Here a = 5, r = \(\frac{4}{7}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{5}{1-\frac{4}{7}}=\frac{35}{3}\)

Question 4.
\(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….
Solution:
Let S = \(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….; Here, a = \(\frac{-3}{4}\) and r = \(\frac{-1}{4}\)

S_∞ = \(\frac{a}{1-r}=\frac{-3 / 4}{1-\left(\frac{-1}{4}\right)}=\frac{-3 / 4}{1+\frac{1}{4}}=\frac{-3}{4} \times \frac{4}{5}=\frac{-3}{5}\)

Question 5.
Prove that 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3
Solution:
LHS = 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3^{\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots .\)};
Here a = \(\frac{1}{2}\) and r = \(\frac{1}{2}\)
∴ S_∞ = 3^{\(\left(\frac{a}{1-r}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1-1 / 2}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1 / 2}\right)\)}
= 3¹
= 3 = R.H.S. [using S_∞ = \(\frac{a}{1-r}\)]

Question 6.
Let x = 1 + a + a² + …….. and y = 1 + b + b² + ….. where |a| < 1 and |b| < 1.
Prove that 1 + ab + a²b² + ….. = \(\frac{x y}{x+y-1}\)
Solution:
Here, x = 1 + a + a² + ……………… = \(\frac{1}{1-a}\) [using S_∞ = \(\frac{a}{1-r}\)]

And y = 1 + b + b² + ………… = \(\frac{1}{1-b}\)

LHS = 1 + ab + a²b² + ……. = \(\frac{1}{1-ab}\) …………(1)

RHS = \(\frac{x y}{x+y-1}=\frac{\left(\frac{1}{1-a}\right)\left(\frac{1}{1-b}\right)}{\left(\frac{1}{1-a}\right)+\left(\frac{1}{1-b}\right)-1}=\frac{\frac{1}{(1-a)(1-b)}}{\frac{1-b+1-a-(1-a)(1-b)}{(1-a)(1-b)}}\)

= \(\frac{1}{(1-a)(1-b)} \times \frac{(1-a)(1-b)}{2-a-b-1+a+b-a b}=\frac{1}{1-a b}\) ……. (2)

From (1) and (2), 1 + ab + a²b² + ………… = \(\frac{x y}{x+y-1}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8c

I.

Question 1.
Find the 20^th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.
Solution:
Given 2 × 4 + 4 × 6 + 6 × 8 + … + n terms
Therefore the n^th term, a_n = 2n (2n + 2) = 4n² + 4n
a₂₀ = 4(20)² + 4(20)
= 4(400) + 80
= 1600 + 80
= 1680
Therefore, 1680 is the 20^th term.

Question 2.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Let the sum of n terms of the G.P be 315. It is known that S_n = \(\frac{1}{2}\)
It is given that a = 5 and common ratio r = 2
Hence, 315 = \(\frac{1}{2}\)
⇒ 2ⁿ – 1 = 63
⇒ 2ⁿ = 64 = (2)⁶
⇒ n = 6
Therefore, last term of the G.P. is 6th term
a₆ = ar^6-1 = ar⁵
= 5 × (2)⁵
= 5 × 32
= 160

Question 3.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Solution:
Let a and r be the first term and common ratio of the G.P respectively.
Hence, a = 1 ⇒ a₃ = ar² = r²
⇒ a₅ = ar⁴ = r⁴
Therefore, r² + r⁴ = 90
r² + r⁴ – 90 = 0
⇒ r² = \(\frac{-1 \pm \sqrt{1+360}}{2}\)
= \(\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}\) = -10 or 9
r² = 9 [Taking real roots]
r = ±3
Thus, the common ratio of the G.P is ±3.

II.

Question 1.
The sum of three consecutive terms in a G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution:
Let the three numbers in G.P. be a, ar and ar².
From the given condition, a + ar + ar² = 56
⇒ a(1 + r + r²) = 56 ………….. (1)
Now, (a – 1), (ar – 7), (ar² – 21) forms an A.P.
Therefore, (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)
⇒ ar – a – 6 = ar² – ar – 14
⇒ ar² – 2ar + a = 8
⇒ a(r² + 1 r 2r) = 8
⇒ 7a(r² – 2r + 1) = 56 ………….(2)
From (1) and (2), we get
⇒ 7(r² – 2r + 1) – (1 + r + r²)
⇒ 7r² – 14r + 7 – 1 – r – r² = 0
⇒ 6r² -15r + 6 = 0
⇒ 2r² – 5r + 2 = 0
⇒ 2r² -4r – r + 2 = 0
⇒ 2r(r – 2) – 1 (r – 2) = 0
⇒ (2r – 1) (r – 2) = 0
⇒ r = \(\frac{1}{2}\), 2

Case I :
Substituting r = 2 in (1), we obtain a(1 + 2 + 22) = 56
⇒ 7a = 56
⇒ a = 8
Hence, the three numbers in G.P are 8, 16 and 32.

Case II :
Substituting r = \(\frac{1}{2}\) in (1), we obtain a(1 + \(\frac{1}{2}\) + \(\left(\frac{1}{2}\right)^2\)) = 56
⇒ \(\frac{1}{2}\)a = 56
a = 32
Hence, the three numbers in G.P are 32, 16 and 8.
Therefore, from case (i) and case (ii), the three required numbers are 8, 16 and 32.

Question 2.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Solution:
Let the G.P be T₁, T₂, T₃, T₄, … T_2n. According to the question,
⇒ T₁ + T₂ + T₃ + T₄ + … + T_2n = 5[T₁ + T₃ + T₅ +… + T_2n-1]
⇒ T₁ + T₂ + T₃ + T₄ +… + T_2n – 5 [T₁ + T₃ + T₅ +… + T_2n-1] = 0
⇒ T₂ + T₄ + ………. + T_2n = 4[T₁ + T₃ + ………. + T_2n-1]
Let the G.P be a, ar, ar²,
\(\frac{\operatorname{ar}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}=\frac{4 \times \mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
⇒ ar = 4a
⇒ r = 4
Thus, the common ratio of the G.P is 4.

Question 3.
If \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\) (x ≠ 0), then show that a, b, c and d are in G.P.
Solution:
It is given that = \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\) (x ≠ 0)
⇒ \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}\) = (a + bx) (b – cx) = (b + cx) (a – bx)
⇒ ab – acx + b²x – bcx² = ab – b²x + acx — bcx²
⇒ 2b²x = 2acx
⇒ b² = ac
⇒ \(\frac{b}{a}=\frac{c}{b}\) (1)
Also, ⇒ \(\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\)
⇒ (b + cx)(c – dx) = (b – cx)(c + dx)
⇒ bc – bdx + c²x – cdx² = bc + bdx – c²x – cdx²
⇒ 2c²x = 2bdx
⇒ c² = bd
⇒ \(\frac{c}{b}=\frac{d}{c}\) …(2)
From Equations (1) and (2), we obtain \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
Thus, a, b, c and d are in G.P.

Question 4.
Let S be the sum, P be the product and R be the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ.
Solution:
Let the G.P be a, ar, ar², ar³ … ar^n-1
According to the given information; Sum of the n terms, S = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
Product of the n terms P = aⁿ × r^{1 + 2 + 3 + …. n-1}
= aⁿr^{\(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)} [∵ Sum of first n natural numbers = n\(\frac{(\mathrm{n}+1)}{2}\)

Sum of reciprocals of n terms R = \(\frac{1}{2}\)
\(\frac{r^{n-1}+r^{n-2}+\ldots+r+1}{a r^{n-1}}=\frac{1\left(r^n-1\right)}{(r-1)} \times \frac{1}{a r^{n-1}}=\frac{r^n-1}{a r^{n-1}(r-1)}\) [∵ 1, r, ……….. r^n-1 forms a G.P]
Therefore, P²Rⁿ = a²ⁿr^n(n-1)\(\frac{\left(r^n-1\right)^n}{a^n r^{n(n-1)}(r-1)^n}=\frac{a^n\left(r^n-1\right)^n}{(r-1)^n}=\left[\frac{a\left(r^n-1\right)}{(r-1)}\right]^n\) = Sⁿ
Hence, P²Rⁿ = Sⁿ proved.

Question 5.
If a, b, c, d are in G.P., prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
Solution:
It is given that a, b, c, d are in G.P.
Therefore, b² = ac ………. (1)
c² = bd … (2)
ad = bc … (3)
We need to prove (an + bn), (bn + cn), (cn + dn) are in G.P i.e.,
(bⁿ + cⁿ)² = (aⁿ + bⁿ) . (cn + dn)
Consider (bⁿ + cⁿ)² = b²n + 2bncn + c²n
= (b²)ⁿ + 2bⁿcⁿ + (c²)ⁿ = (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ [Using (1) and (2)]
= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ) = (aⁿ + bⁿ) (cⁿ + dⁿ)
Hence, (bⁿ + cⁿ)² = (aⁿ + bⁿ) . (cⁿ + dⁿ)
Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Question 6.
If a and b are the roots of x² – 3x + p * 0 and c, d are the roots of x² – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15.
Solution:
It is given that a and b are the roots of x² – 3x + p = 0
Therefore, a + b = 3, ab = p
Also, c and d are the roots of x² – 12x + q = 0; c + d = 12, cd = q
It is given that a, b, c, d form a G.P.
Let a = x, b = xr, c = xr², d = xr³
From (1) and (2), we obtain, ⇒ x + xr = 3
⇒ x(1 + r) = 3
⇒ xr² + xr³ = 12
⇒ xr² (1 + r) = 12 … (4)

On dividing (4) by (3), we obtain \(\frac{\mathrm{xr}^2(1+\mathrm{r})}{\mathrm{x}(1+\mathrm{r})}=\frac{12}{3}\)
⇒ r² = 4 ⇒ r = ±2

Case I:
When r = 2, x = 1 ⇒ ab = x²r = 2 & cd = x²r⁵ = 32
Therefore, ⇒ \(\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\)
⇒ (q + p) : (q – p) = 17 : 15

Case II: when r = -2, x = -3
⇒ ab = x²r = -18 & cd = x²r⁵ = -288
Therefore, ⇒ \(\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\)
Thus, (q + p) : (q – p) = 17 : 15 is proved.

Question 7.
The ratio of the A.M and G.M. of two positive numbers a and h, is m : n. Show that a : b = (m +\(\sqrt{m^2-n^2}\)):(m – \(\sqrt{m^2-n^2}\)).
Solution:
Let the two numbers be a and b. A.M = \(\frac{a+b}{2}\), G.M = \(\sqrt{a b}\)
Adding (1) and (2), we obtain \(\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \Rightarrow \frac{(a+b)^2}{4 a b}=\frac{m^2}{n^2}\)
⇒ (a + b)² = \(\frac{4 \mathrm{abm}^2}{\mathrm{n}^2}\)
⇒ (a + b) = \(\frac{2 \mathrm{~m} \sqrt{\mathrm{ab}}}{\mathrm{n}}\) .. (1)

Using this in the identity (a – b)² = (a + b)² – 4ab, we obtain

Substituting the value of a in (1), we obtain

III.

Question 1.
Find the sum of the following series upto n terms.
(i) 5 + 55 + 555 + …
Solution:
Given that 5 + 55 + 555 + …
Let S_n = 5 + 55 + 555 + … n terms
S_n = \(\frac{5}{9}\)[9 + 99 + 999 + …n terms] = \(\frac{5}{9}\) (10 – 1) + (10² – 1) + (10³ – 1) + …nterms
= \(\frac{5}{9}\)[(10 + 10² + 10³ + ……………. n terms) – (1 + 1 + ….. n terms)]
= \(\frac{5}{9}\)[\(\frac{10\left(10^{\mathrm{n}}-1\right)}{10-1}\) – n]

= \(\frac{5}{9}\)[\(\frac{10\left(10^{\mathrm{n}}-1\right)}{9}\) – n]

= \(\frac{50}{81}\)(10ⁿ – 1) – \(\frac{5n}{9}\)

(ii) .6 + .66 + .666 + …
Solution:
S_n = .6 + .66 + .666 + … n terms
= 6[.1 + .11 + .111 + …… upto nterms] = \(\frac{6}{9}\)[0.9 + 0.99 + 0.999 + upto n terms]

Therefore, the sum of n terms of the given series is \(\frac{2}{3}\) n – \(\frac{2}{27}\)(1 – 10^-n)

Question 2.
If f is a function satisfying f(x + y) = f(x) . f(y) for all x, y ∈ N such that f(1) = 3 and \(\sum_{x=1}^n\) f(x) = 120. And the value of n.
Solution:
f(x + y) = f(x) . f(y) for all x, y ∈ N and f(1) = 3
Taking x = y = 1 in the given function, we obtain f(x + y) = f(x) . f(y) ⇒ f(l + 1) = f(1) f(1) f(2) = 3 x 3 = 9
Similarly, f(x + y) = f(x) . f(y) ⇒ f(1 + 2) = f(1) f(2) ⇒ f(3) = 3 x 9 = 27
Also, f(x + y) = f(x) . f(y) ⇒ f(1 + 3) = f(1) f(3) ⇒ f(4) = 3 x 27 = 81
Therefore, f(1), f(2), f(3) i.e., 3, 9, 27, 81 forms a G.P. with both the first term and common ratio equal to 3.
It is known that S_n = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\). It is given that \(\sum_{x=1}^n\) f(x) = 120

⇒ 120 = \(\frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}\)
⇒ 120 = \(\frac{3}{2}\)(3ⁿ – 1)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 3⁴
⇒ n = 4
Thus, the value of n = 4.

Question 3.
A fanner buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual installments of ₹ 500 plus 12% interest on the unpaid amount. How much will the tractor cost him ?
Solution:
Given, the farmer pays ₹ 6000 in cash.
So, the unpaid amount = ₹ 12000 – ₹ 6000 = ₹ 6000
From the question, the interest paid annually will be 12% of 6000, 12% of 5500, 12% of 5000, … , 12% of 500
Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500 = 12% of (6000 + 5500 + 5000 + … + 500)
= 12% of (500 + 1000 + 1500 + … + 6000)
It’s seen that the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both are equal to 500.
Let’s take the number of terms of the A.P to be n.
So, 6000 = 500 + (n – 1) 500
⇒ 1 + (n – 1) = 12
⇒ n = 12
Now, The sum of the series in A.P = \(\frac{12}{2}\) [2(500) + (12 – 1) (500)]
= 6[1000 + 5500]
= 6(6500)
= 39000
Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 +
= 12% of 39000 = ₹4680
Therefore, the tractor will cost the farmer = (₹ 12000 + ₹ 4680) = ₹ 16680

Question 4.
Shamshad Ali buys a scooter for ₹ 22000. He pays ₹ 4000 cash and agrees to pay the balance in annual installment of? 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him ?
Solution:
It is given that Shamshad Ali buys a scooter for ₹ 22000 and pays ₹ 4000 in cash. Hence, unpaid amount ₹ = 22000 – 4000 = 18000
According to the given condition, the interest paid annually will be (10% of 18000), (10% of 17000), (10% of 16000), … (10% of 1000)
Thus, total interest to be paid = (10% of 18000) + (10% of 17000) + … + (10% of 1000) = 10% of (18000 + 17000 + … + 1000) = 10% of (1000 + 2000 + … + 18000)
Now, the series 1000 + 2000 + … + 18000 forms an A.P. with a and d both equal to 1000.
Let the number of terms of the A.P. be n.
⇒ a_n = a + (n – 1) d
⇒ 18000 = 1000 + (n – 1) (1000)
⇒ 18000 – 1000 = 1000 (n – 1)
⇒ n – 1 = \(\frac{17000}{1000}\)
⇒ n = 17 + 1
⇒ n = 18
Now, total cost of scooter in ₹ = 22000 + 17100 = 39100
Thus, the scooter will cost him ₹ 39100.

Question 5.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter, find the amount spent on the postage when the 8th set of letters is mailed.
Solution:
The number of letters mailed forms a G.P : 4, 4², ……….. , 4⁸; Here a = 4, n = 8
Hence, S = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
S8 = \(\frac{4\left(4^8-1\right)}{4-1}=\frac{4(65536-1)}{3}=\frac{4 \times 65535}{3}\)
= 4 × 21845
= 87380
It is given that the cost to mail one letter is 50 paisa.
Therefore, cost in ₹ for mailing 87380 letters = \(\frac{50}{100}\) × 87380 = 43690
Thus, ₹ 43690 is the amount on the postage when 8th set of letters is mailed.

Question 6.
A man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount, also calculate the total amount after 20 years.
Solution:
It is given that the man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Hence, annual interest received in ₹ = \(\frac{50}{100}\) × 10000 = 500
Amount in ₹ at the end of first year = 10000 + 500 = 10500
Amount in ₹ at the end of second year = 10500 + 500 = 11000
Amount in ₹ at the end of third year = 11000 + 500 = 11500
Therefore, 10000, 10500, 11000, … is an A.P.
Hence, the amount in the 15^th year,
a₁₅ = 10000 + (15 – 1) (500) = 10000 + 7000 = 17000
Amount after 20 years
a₂₁ = 10000 + (21 – 1) (500) = 10000 + 10000 = 20000
Thus, the amount in 15th year is ₹ 17000 and after 20 years amount is ₹ 20000.

Question 7.
A manufacturer reckons that the value of a machine, which costs him ₹ 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Solution:
Cost of a machine is ₹ 15625. Machine depreciates by 20% every year.
∴ Its value after every year is 80% of the original cost i.e., 4/5 of the original cost.
Value at the end of 5 years =

= 15628 × \(\left(\frac{4}{5}\right)^5\)
= 5 × 1024
= 5120
Thus, value of the machine at the end of 5 years is ₹ 5120.

Question 8.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution:
Let x be the number of days in which 150 workers finish the work.
According to the given information, 150x = 150 + 146 + 142 + … (x + 8) terms
Here, 150 + 146 + 142 + … (x + 8) terms forms an A.P. with a = 150, d = – 4, n = (x + 8)
Therefore 150x = \(\frac{(x+8)}{2}\)[2(150) + (x + 8 -1)(-4)]
⇒ 150x – (x + 8) [150 + (x + 7) (-2)]
⇒ 150x = (x + 8) (150 – 2x – 14)
⇒ 150x = (x + 8) (136 – 2x)
⇒ 75x – (x + 8) (68 – x)
⇒ 75x – 68x – x² + 544 – 8x
⇒ x² + 15x – 544 = 0
⇒ x² + 32x – 17x – 544 = 0
⇒ x(x + 32) – 17(x + 32) = 0
⇒ (x – 17) (x + 32) = 0
⇒ x = 17, – 32
However, x cannot be negative, hence, x = 17
∴ The number of days in which the work was completed is (17 + 8) = 25 days. Thus, the work was completed in 25 days.

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8e

I. Find the sum to n terms of each of the series in Exercises 1 to 7.

Question 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Solution:
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n terms
Hence, a_n = n(n + 1)
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k(k + l) = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+1}{3}+1\right)\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+4}{3}\right)\)

= \(\frac{\mathrm{n}}{3}\)(n + 1)(n + 2)

Question 2.
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution:
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n terms
Hence, a_n = n(n + 1) (n + 2) = (n² + n) (n + 2) = n³ + 3n² + 2n
Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\)(k³ + 3k² + 2k) = \(\sum_{k=1}^n\) k³ + 3 \(\sum_{k=1}^n\) k² + 2 \(\sum_{k=1}^n\) k

Question 3.
3 × 1² + 5 × 2² + 7 × 3² + …
Solution:
The given series is 3 × 1² + 5 × 2² + 7 × 3² + …. n terms
Hence, a_n = (2n + 1) n² = 2n³ + n²
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (2k³ + k²) = 2\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k³ + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k²

= 2\(\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\) + \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{2}+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\mathrm{n}(\mathrm{n}+1)+\frac{2 \mathrm{n}+1}{3}\right)\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+3 n+2 n+1}{3}\right]\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+5 n+1}{3}\right]\)

= \(\frac{n}{6}\)(n + 1)(3n + 5n + 1)

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ………..
Solution:
The given series is \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ……….. n terms
Hence, a_n = \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
∴ a₁ = \(\frac{1}{1}-\frac{1}{2}\), a₂ = \(\frac{1}{2}-\frac{1}{3}\), a₃ = \(\frac{1}{3}-\frac{1}{4}\), …………….., a_n = \(\frac{1}{n}-\frac{1}{n+1}\),

Adding the above terms column wise, we obtain
a₂ + a₂ + …… + a_n = \(\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{\mathrm{n}}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\mathrm{n}+1}\right]\)

Thus, S_n = 1 – \(\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Question 5.
5² + 6² + 7² + ………. +20²
Solution:
The given series is 5² + 6² + 7² + ………. +20²
Hence, a_n = (n + 4)² = n² + 8n + 16

Therefore, S_n = \(\sum_{k=1}^n\)a_k = \(\sum_{k=1}^n\)(k² + 8k + 16)

S_n = \(\sum_{k=1}^n\)k² + 8\(\sum_{k=1}^n\)k + \(\sum_{k=1}^n\) 16

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{8 \mathrm{n}(\mathrm{n}+1)}{2}\) + 16n

Since, 16^th term is (16 + 4)² = (20)²

Then, S_n = \(\frac{16(16+1)(2(16)+1)}{6}+\frac{8(16)(16+1)}{2}\)+ 16(16)

= \(\frac{(16)(17)(33)}{6}+\frac{(128)(17)}{2}\) + 256

= 1496 + 1088 + 256 = 2840

Thus, 5² + 6² + 7² + ………. +20² = 2840

Question 6.
3 × 8 + 6 × 11 + 9 × 14 + …………..
Solution:
The given series is 3 × 8 + 6 × 11 + 9 × 14 + ………….. n terms
Hence, a_n = (n^th term of 3, 6, 9, ………..) × (n^th term of 8, 11, 14 ……)
= (3n)(3n + 5)
= 9n² + 15n

Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\) (9k² + 15k)

= 9 \(\sum_{k=1}^n\) k² + 15\(\sum_{k=1}^n\)k

= 9 × \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) + 15 × \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{2}+\frac{15 \mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 1 + 5)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 6)

= 3n(n + 1)(n + 3)

Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………..
Solution:
The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + ……….. n terms

Hence, a_n = (1² + 2² + 3² + …… + n²) = \(\frac{n}{2}\)(n + 1)(2n + 1)
= \(\frac{n}{6}\) (2n² + 3n + 1)
= \(\frac{1}{3}\)n³ + \(\frac{1}{2}\)n² + \(\frac{1}{6}\)n

Therefore,

II. Find the sum to n terms of the series in Exercises 1 to 3 whose nth terms is given by

Question 1.
n(n + 1) (n + 4).
Solution:
The given n* term is a_n = n(n + 1) (n + 4)
Hence, a_n = n(n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n
Therefore, S_n = \(\sum_{k=1}^n\) a_k = \(\sum_{k=1}^n\)k³ + 5 \(\sum_{k=1}^n\) k² + 4 \(\sum_{k=1}^n\) k

Question 2.
n² + 2n
Solution:
The given n^th term is an = n² + 2ⁿ
Hence, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + 2^k

= \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)2^k …………..(1)
Consider \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = 2¹ + 2² + 2³ +…

The above series 2¹ + 2² + 2³ + … is a G.P with both the first term and common ratio equal to 2.
\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = \(\frac{1}{2}\) = 2(2ⁿ – 1) …………..(2)

From (1) and (2) we obtain
S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k
= \(\frac{n}{6}\)(n + 1)(2n + 1) + 2(2ⁿ – 1)

Question 3.
(2n – 1)²
Solution:
The given n^th term is a_n = (2n – 1)²
Hence, a_n = (2n – 1)² = 4n² – 4n + 1

Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (4k² – 4k + 1)

= 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k2 – 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 1

= \(\frac{4 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{4 \mathrm{n}(\mathrm{n}+1)}{2}\) + n

= \(\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\) – 2n(n + 1) + n

= n[\(\frac{2\left(2 n^2+3 n+1\right)}{3}\) – 2(n + 1) + 1]

= n\(\left[\frac{4 n^2+6 n+2-6 n-6+3}{3}\right]\) = n\(\left[\frac{4 \mathrm{n}^2-1}{3}\right]\)

= \(\frac{\mathrm{n}}{3}\) (2n + 1)(2n – 1)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7a

I. Expand each of the expressions in Exercises 1 to 8.

Question 1.
(1 – 2x)⁵
Solution:
By using Binomial theorem, the expression (1 – 2x)⁵ can be expanded as
⁵C₀(1)⁵(-2x)⁰ + ⁵C₁(1)⁴(-2x)¹ + ⁵C₂ (1)³ (-2x)² + ⁵C₃(1)²(-2x)³ + ⁵C₄(1)¹ (-2x)⁴ + ⁵C₅(1)° (-2x)⁵
= 1 – 5(2x) + 10(4x²) – 10(8x³) + 5(16x⁴) – (32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵.

Question 2.
\(\left(\frac{2}{x}-\frac{x}{2}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\) can be expanded as

Question 3.
(2x – 3)⁶
Solution:
By using Binomial theorem, the expression (2x – 3)⁶ can be expanded as ⁶C₀ (2x)⁶ (-3)° + ⁶C₁ (2x)⁵ (-3)¹ + ⁶C₂ (2x)⁴ (-3)² + ⁶C₃ (2x)³ (-3)³ + ⁶C₄ (2x)² (-3)⁴ + ⁶C₅ (2x)¹ (-3)⁵ + ⁶C₆ (2x)⁰ (-3)⁶
= 64x⁶ – 6(32x⁵) (3) + 15(16x⁴)(9) – 20(8x³)(27) + 15(4x²)(81) – 6(2x)(243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Question 4.
\(\left(\frac{x}{3}+\frac{1}{x}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^5\) can be expanded as

Question 5.
\(\left(x+\frac{1}{x}\right)^6\)
Solution:
By using Binomial theorem, the expression \(\left(x+\frac{1}{x}\right)^6\) can be expanded as

Question 6.
(4x + 5y)⁷
Solution:
By using Binomial theorem,
(4x + 5y)⁷ = ⁷C₀ (4x)⁷ (5y)⁰ + ⁷C₁ (4x)⁶ (5y)¹ + ⁷C₂ (4x)⁵ (5y)² + ⁷C₃ (4x)⁴ (5y)³ + ⁷C₄ (4x)³ (5y)⁴ + ⁷C₅ (4x)² (5y)⁵ + ⁷C₆ (4x)¹ (5y)⁶ + ⁷C₇ (4x)° (5y)⁷
= \(\sum_{r=0}^7\) ⁷C_r (4x)^7-r (5y)^r

Question 7.
\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
Solution:
By using Binomial theorem, we have

Question 8.
\(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:

Question 9.
Write down and simplify.
i) 6^th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
Given that \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is
T_r+1 = ⁹C_r \(\left(\frac{2 x}{3}\right)^{9-r}\left(\frac{3 y}{2}\right)^r\)
If r = 5
T₆ = ⁹C₅\(\frac{1}{2}\)
= ⁹C₅\(\frac{1}{2}\)
= ⁹C₄\(\frac{1}{2}\) x⁴y⁵
= \(\frac{1}{2}\). x⁴y⁵
= 189 x⁴y⁵

ii) 7^th term in (3x – 4y)¹⁰.
Solution:
Given that (3x – 4y)¹⁰
The general term in (3x – 4y)¹⁰ is T_r+1 = (-1)^{r 10}C_r (3x)^10-r (4y)^r
If r = 6 ⇒ T₇ = (-1)^{6 10}C₆ (3x)⁴ (4y)⁶
= ₂C₄3₄4₆x₄y₆
= \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\). 3₄.4₆. x₄. y₆
= 280 (12)₅x₄y₆ (∵ _nC_r = _nC_n-r)

iii) 10^th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
Given that \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
The general term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is T_r+1 = (-1)^{r 14}C_r \(\left(\frac{3 \mathrm{p}}{4}\right)^{14-\mathrm{r}}\) (5q)^r
If r = 9
T₁₀ = (-1)^{9 14}C₉\(\left(\frac{3 p}{4}\right)^5\) (5q)⁹
= – ¹⁴C₅\(\left(\frac{3}{4}\right)^5\) 5⁹.p⁵.q⁹ (∵ _nC_r = _nC_n-r)
= \(-\frac{14 \times 13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4 \times 5} \cdot \frac{3^5}{4^5}\).5⁹.p⁵.q⁹
= \(\frac{-(2002) 3^5 \cdot 5^9}{4^5}\)p⁵.q⁹

iv) r^th term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 8)
Solution:
Given that \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is T_r+1 = ⁸C_r \(\left(\frac{3 a}{5}\right)^{8-r}\left(\frac{5 b}{7}\right)^r\)
Replace r by r – 1, then
T_r = ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{8-r+1}\left(\frac{5 b}{7}\right)^{r-1}\)
= ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{9-r}\left(\frac{5 b}{7}\right)^{r-1}\) (1 ≤ r ≤ 8)

v) Middle term in \(\left(3-\frac{x^3}{6}\right)^7\)
Solution:
Given that \(\left(3-\frac{x^3}{6}\right)^7\)

The expression contains 8 terms. Thus T₄ and T₅ are middle terms.
T₄ = T₃₊₁ = ⁷C₃ (3)^7-3\(\left(-\frac{x^3}{6}\right)^3=\frac{35 \times 3^4(-x)^9}{6^3}=-\frac{105}{8}\)x⁹.

T₅ = T₄₊₁ = ⁷C₄ (3)^7-4\(\left(-\frac{x^3}{6}\right)^4=\frac{35 \times 3^3 x^{12}}{6^4}=\frac{35}{48}\)x¹²

Thus, the middle terms in the expansion of \(\left(3-\frac{x^3}{6}\right)^7\) are \(-\frac{105}{8}\) x⁹ and \(\frac{35}{48}\)x¹²

vi) Middle term in \(\left(\frac{x}{3}+9 y\right)^{10}\)
Solution:
Given that \(\left(\frac{x}{3}+9 y\right)^{10}\)
The number of terms in the given expansion is 11. Thus T₆ is the middle term.
T₆ = T₅₊₁ = ¹⁰C₅ \(\frac{1}{2}\) (9y)⁵
= \(\frac{1}{2}\) 3¹⁰.x⁵y⁵
= 252 × 3⁵ x⁵.y⁵
= 61236 x⁵y⁵

vii) 4th term in (x – 2y)¹²
Solution:
Given that (x – 2y)¹²
General term in the expansion of (x + y)ⁿ is Tr+ x = nCr Xn_r (Y)r
Here, X = x, Y = -2y, and n = 12; T_r+1 = ¹²C_r (x)^12-r (-2y)^r
Putting r = 3, ,
T₃₊₁ = ¹²C₃x^12-3 (-2y)³
T₄ = \(\frac{12 \times 11 \times 10}{6}\) x⁹(-2)³y³
= 2 × 11 × 10x⁹(-8)y³
= 220 × (-8)x⁹y³
= -1760 x⁹y³

viii) 13^th term in \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\), x ≠ 0
Solution:
The general term in the expansion of (X + Y)ⁿ is T_r+1 = ⁿC_r X^n-rY^r

Here X = 9x, Y = –\(\frac{1}{3 \sqrt{x}}\), and n = 18

The General term in the expansion of \(\frac{1}{2}\) is T_r = ¹⁸C_r (9x)^18-r\(\left(\frac{-1}{3 \sqrt{x}}\right)^r\)
Putting r = 12
T₁₂₊₁ = ¹⁸C₁₂ (9x)⁶ \(\left(\frac{-1}{3 \sqrt{x}}\right)^{12}\)
= ¹⁸C₁₂ 9⁶ × x⁶\(\frac{1}{3^{12}(\sqrt{x})^{12}}\) (∵ _nC_r = _nC_n-r)
= \(\frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \times \frac{3^{12} \times x^6}{3^{12} \times x^6}\)
= 18564

Question 10.
Find the number of terms in the expansion of
i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

ii) (3p + 4q)¹⁴
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in (3p + 4q)¹⁴ are 14 + 1 = 15

II. Using binomial theorem, evaluate each of the following.

Question 1.
(96)³
Solution:
96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied, it can be written that, 96 = 100 – 4
Therefore, (96)³ = (100 – 4)³
= ³C₀ (100)³ (-4)⁰ + ³C₁ (100)² (-4)¹ + ³C₂ (100)¹ (-4)² + ³C₃ (100)° (-4)³
= (100)³ – 3(100)²(4) + 3(100)(4)² – (4)³
= 1000000 – 120000 + 4800 – 64
= 884736

Question 2.
(102)⁵
Solution:
102 can be expressed as the sum or difference of two numbers whose powers are
easier to calculate and then, Binomial theorem can be applied.
It can be written that, 102 = 100 + 2.
Therefore, (102)⁵ = (100 + 2)⁵
= ⁵C₀ (100)⁵ (2)° + ⁵C₁ (100)⁴ (2)¹ + ⁵C₂ (100)³ (2)² + ⁵C₃ (100)² (2)³ + ⁵C₄ (100)¹ (2)⁴ + ⁵C₅ (100)° (2)⁵
= (100)⁵ + 5(100)⁴ (2) + 10(100)³(2)² + 10(100)²(2)³ + 5(100) × (2)⁴ + (2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

Question 3.
(101)⁴
Solution:
101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that, 101 = 100+ 1.
(101)⁴ = (100 + 1)⁴ = ⁴C₀ (100)⁴(1)°+ ⁴C₁ (100)³ (1)¹ + ⁴C₂ (100)² (1)² + ⁴C₃ (100)¹ (1)³ + ⁴C₄ (100)° (1)⁴
= (100)⁴ + 4(100)³ + 6(100)² + 4(100)¹ + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

Question 4.
(99)⁵
Solution:
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that 99 = 100 – 1
=> (99)⁵ = (100 – 1)⁵
= ⁵C₀(100)⁵(-1)° + ⁵C₁(100)⁴(-1)¹ + ⁵C₂(100)³(-1)² + ⁵C₃(100)²(-1)³ + ⁵C₄(100)¹(-1)⁴ + ⁵C₅(100)°(-1)⁵
= (100)⁵ – 5(100)⁴ + 10(100)³ – 10(100)² + 5(100) – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500100001
= 9509900499

Question 5.
Using Binomial theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ 0r 1000.
Solution:
By splitting 1.1 and then applying Binomial theorem, the first few terms of (1.1)10000 can be obtained as (1.1)¹⁰⁰⁰⁰
= (1 + 0.1)¹⁰⁰⁰⁰
= ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1)¹ + other positive terms
= 1 + 10000 × 0.1 + other positive terms
= 1001 + other positive terms > 1000
Hence, (1.1)¹⁰⁰⁰⁰ > 1000

Question 6.
Find (a + b)⁴ – (a – b)⁴, hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴
Solution:
Using Binomial theorem, the expression, (a + b)4 – (a – b)4, can be expanded as (a + b)⁴ – (a – b)⁴
= ⁴C₀a⁴b° + ⁴C₁b¹ + ⁴C₂a²b² + ⁴C₃b³ + ⁴C₄a°b⁴ – [⁴C₀a⁴b° – ⁴C₁a³b¹ + ⁴C₂a²b² – ⁴C₃a¹b³ + ⁴C₄a°b⁴]
= 2[⁴C₁a³b¹ + ⁴C₃a¹b³]
= 2[4a³b + 4ab³]
= 8ab(a² + b²)

Now putting a = √3 and b = √2 , we have
(√3 + √2)⁴ – (√3 – √2)⁴ = 8(√3)(√2) [(√3)² +(√2)²]
= 8√6(3 + 2)
= 40√6

Question 7.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1) + (√2 – 1)
Solution:
Using Binomial theorem, the expression, (x + 1)⁶ + (x – 1)⁶, can be expanded as (x + 1)⁶ + (x – 1)⁶
= ⁶C₀x⁶1⁰ + ⁶C₁⁵1¹ + ⁶C₂x⁴1² + ⁶C₃x³1³ + ⁶C₄x²1⁴ + ⁶C₅x¹1⁵ + ⁶C₆x°1⁶ + ⁶C₀x⁶1°- ⁶C₁x⁵1¹ + ⁶C₂x⁴1² – ⁶C₃x³1³ + ⁶C₄x²1⁴ – ⁶C₅x¹1⁵ + ⁶C₆x°1⁶
= 2[⁶C₀x⁶ + ⁶C₂x⁴ + ⁶C₄x² + ⁶C₆x⁰]
= 2(x⁶ + 15x⁴ + 15x² + 1)
Now putting x = √2 , we have (√2)⁶ + (√2)⁶
= 2[(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1]
= 2(8 + 60 + 30 + 1)
= 198

Question 8.
Show that 9n +1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Solution:
Using Binomial theorem,
(1 + x)ⁿ⁺¹ = ⁿ⁺¹C₀ x° + ⁿ⁺¹C₁ x¹ + ⁿ⁺¹C₂ x² + …………. + ⁿ⁺¹C_n+1 xⁿ⁺¹
Now putting x = 8, we have
(1+8)ⁿ⁺¹ = ⁿ⁺¹C₀ 8° + ⁿ⁺¹C₁ 8¹ + ⁿ⁺¹C₂ 8² + ⁿ⁺¹C_n+1 8ⁿ⁺¹
⇒ 9ⁿ⁺¹ = 1 + (n + 1)(8) + 8²[ⁿ⁺¹C₂ + …….. + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 1 – 8(n+1) = 64[ⁿ⁺¹C₂ + + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 8n – 9 = 64 [ⁿ⁺¹C₂ + …. + ⁿ⁺¹C_n+1 8^n-1]
Now, RHS is divisible by 64 for all positive integers n. So, LHS is also divisible by 64. Hence, 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive interger.

Question 9.
Prove that \(\sum_{r=0}^n\)3^rnC_r = 4ⁿ.
Solution:
LHS = \(\sum_{r=0}^n\)3^rnC_r = 3⁰ⁿC₀ + 3¹ⁿC₁ + 3²ⁿC₂ + …………… 3^{n-1 n}C_n-1 + ⁿC_n+1. 3ⁿ

= ⁿC₀ 3° + ⁿC₁ 3¹ + ⁿC₂ 3² + …. + 3^n-1 + ⁿC_n.3ⁿ
= (1 + 3)ⁿ = 4ⁿ = RHS

Question 10.
Using the binomial theorem, prove that 50” – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
Given 50ⁿ – 49n – 1 = (49 + 1)ⁿ – 49n – 1
= ⁿC₀(49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n 2(49)² + ⁿC_n-1(49) + ⁿC_n(49)° – 49n – 1
= (49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n-2(49)² + (n) (49) + 1 – 49n – 1
= (49)² [(49)^n-2 + ⁿC₂ (49)^n-3 + ⁿC₂(49)^n-4 + … + ⁿC_n-2]
= (49)² [a positive integer]
∴ 50ⁿ – 49n – 1 is divisible by 492 for all positive integers of n.

Practice AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Sequences and Series MCQ

Question 1.
Which of the following is finite sequence ?
1) 48, 24, 12, ………..
2) 1, 2, 3, …….
3) 2, 4, 6, 8, 10 ………..
4) 2, 3, 5, 7, 11, 13, …….
Answer:
3) 2, 4, 6, 8, 10 ………..

Explanation:
2, 4, 6, 8, 10 is a finite sequence.

Question 2.
Which of the following relation gives Fibonacci sequence ?
1) a_n = a_n-1 + a_n-2, n > 2
2) a_n-1 = a_n + a_n-2, n > 2
3) a_n-2 = a_n + a_n-1, n> 2
4) a_n = a_n+1 + a_n-2, n > 2
Answer:
1) a_n = a_n-1 + a_n-2, n > 2

Explanation:
Relation of fibonacci sequence is a_n = a_n-1 + a_n-2, n > 2

Question 3.
The first term of Fibonacci sequence is
1) 0
2) 1
3) 2
4) 3
Answer:
2) 1

Explanation:
The first term of fibonacci sequence is “1”.

Question 4.
If a_n = 4n + 6, then 15^th term of the sequence is
1) 6
2) 10
3) 60
4) 66
Answer:
4) 66

Explanation:
a_n = 4n + 6
a₁₅ = 4(15) + 6
= 60 + 6 = 66

Question 5.
A series can also be denoted by the symbol …………
1) πa_n
2) Σa_n
3) Φa_n
4) θa_n
Answer:
2) Σa_n

Explanation:
Series denoted by Σa_n

Question 6.
The sum of first five terms of series 2 + 4 + 6 + ….. is
1) 14
2) 16
3) 20
4) 30
Answer:
4) 30

Explanation:
Given series 2 + + 6 + ………..
Sum of even number S_n = n(n + 1)
S₅ = 5(5 + 1) = 5.(6) = 30

Question 7.
\(\sum_{n=1}^4\) 2n + 3 = ……….
1) 5
2) 12
3) 21
4) 32
Answer:
4) 32

Explanation:
\(\sum_{n=1}^4\) 2n + 3, If n = 1, 2, 3, 4; Series is 5, 7, 9, 11 …… ; S₄ = 5 + 7 + 9 + 11 = 32

Question 8.
If a_n+1 = a_n. r then the sequence is called
1) Arithmetic progression
2) Geometric progression
3) Harmonic progression
4) Special progression
Answer:
2) Geometric progression

Explanation:
a_n+1 = a_n. r is G.P.

Question 9.
The nth term of G.P. is ( 3 )
1) a_n = a + (n – 1) d
2) a_n = a + nd
3) a_n = a . r^n-1
4) a_n = arⁿ
Answer:
3) a_n = a . r^n-1

Explanation:
n^th term of G.P. is a_n = a . r^n-1

Question 10.
If r = 1 in G.P. then the sum of first n terms is
1) na
2) a/n
3) (n – 1) a
4) (n + 1) a
Answer:
1) na

Explanation:
If r = 1, the sum of first n terms = na

Question 11.
If 4, 8, 16, 32 are in G.P. then the sum upto 5th term is
1) 16
2) 64
3) 128
4) 124
Answer:
4) 124

Explanation:
4, 8, 16, 32, are in G.P; a = 4, r = 2
S₅ = \(\frac{4\left(2^5-1\right)}{2-1}=\frac{4(32-1)}{1}\) = 4(31) = 124

Question 12.
The Sum of series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms is
1) \(\frac{63}{32}\)
2) \(\frac{32}{63}\)
3) \(\frac{26}{53}\)
4) \(\frac{53}{26}\)
Answer:
1) \(\frac{63}{32}\)

Explanation:
Series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms
Sum of 6 terms = \(\frac{\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}\)
= 2(1 – \(\frac{1}{64}\))
= 2 – \(\frac{1}{32}=\frac{64-1}{32}=\frac{63}{32}\)

Question 13.
The geometric mean of 3 and 12 is ………….
1) 4
2) 6
3) 9
4) 12
Answer:
2) 6

Explanation:
Geometric mean of 3 and 12 = \(\sqrt{3 \times 12}=\sqrt{36}\)
= 6

Question 14.
If A.M of two numbers is \(\frac{15}{2}\) and their G.M is 6 then the two numbers are
1) 6 and 8
2) 12 and 3
3) 24 and 6
4) 27 and 3
Answer:
2) 12 and 3

Explanation:
Let the two numbers be and b
A.M = \(\frac{15}{2} \Rightarrow \frac{a+b}{2}=\frac{15}{2}\)
⇒ a + b = 15 ……….(1)

G. M = 6 ⇒ \(\sqrt{\mathrm{ab}}\)
⇒ ab = 36

a – b = \(\sqrt{(a+b)^2-4 a b}=\sqrt{225-144}=\sqrt{81}\)
a – b = 9 …(2)
(1) + (2) ⇒ a = 12, b = 3

Question 15.
If A means arithmetic mean and G means geometric mean then which of the following is true ?
1) A > G
2) A ≥ G
3) G > A
4) G ≥ A
Answer:
2) A ≥ G

Explanation:
A ≥ G

Question 16.
The ratio of A.M. and G.M of two positive numbers a and b is 5 : 3. Then the ratio of a to b is ( 1 )
1) 9 : 1
2) 3 : 5
3) 1 : 9
4) 3 : 1
Answer:
\(\frac{A \cdot M}{G \cdot M}=\frac{5}{3} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{5}{3}\)
⇒ 9(a + b)² = 25 (4ab)
⇒ 9a² + 18ab + 9b² = 100 ab
⇒ 9a² – 82ab + 9b² = 0
⇒ (9a – b) (a – 9b) = 0
9a-b = 0
9a = b
\(\frac{a}{b}=\frac{1}{9}\)

a – 9b = 0
a = 9b
\(\frac{a}{b}=\frac{9}{1}\)
∴ a : b = 9 : 1 (∵ a > b)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7b

I.

Question 1.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint : Write aⁿ = (a – b + b)ⁿ and expand]
Solution:
Here, aⁿ = (a – b + b)ⁿ = [(a – b) + b)ⁿ
By using Binomial theorem, the expression [(a – b) + b]ⁿ can be expanded as
ⁿC₀(a- b)ⁿ b⁰ + ⁿC₁(a- b)^n-1 b¹ + ⁿC₂(a- b)^n-2 b² + …. + ⁿC_n-1(a – b)b^n-1 + ⁿC_n(a- b)⁰bⁿ
= (a – b)ⁿ + ⁿC₁(a – b)^n-1 b + ⁿC₂(a – b)^n-2 bⁿ + ………….. + ⁿC_n-1(a – b)b^n-1] + bⁿ
⇒ aⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ⁿC_n-1b^n-1] + bⁿ
⇒ aⁿ – bⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ……. + ⁿC_n-1b^n-1]
now, a – b is a factor of R.H.S, so, a – b is also a factor of L.H.S hence, a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Question 2.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶
Solution:
(√3 + √2)⁶ – (√3 – √2)⁶
= ⁶C₀(√3)⁶ + ⁶C₁(√3)⁵(√2) + ⁶C₂(√3)⁴(√2)² + ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴
+ ⁶C₅(√3)¹(√2)⁵ + ⁶C₆(√2)⁶ – [⁶C₀(√3)6 – ⁶C₁(√3 )⁵(√2) + ⁶C₂(√3)⁴(√2)² – ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴ – ⁶C₅(√3)¹(√2 )⁵ + ⁶C₆(√2)⁶]
= 2[⁶C₁(√3)⁵(√2) + ⁶C₃(√3)³(√2)³ + ⁶C₂(√3)¹(√2)⁵]
= 2[6 × 9√6 + 20 × 6√6 + 6 × 4√5]
= 2[54 + 120 + 24]√6
= 396√6

Question 3.
Find the value of (a² + \(\sqrt{a^2-1}\))⁴ + (a² – \(\sqrt{a^2-1}\))⁴
Solution:

= 2(a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]
= 2[a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

II.

Question 1.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Solution:
0.99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then Binomial theorem can be applied. It can be written that, 0.99 = 1 – 0.01
Therefore, (0.99)⁵ = (1 – 0.01)⁵
= ⁵C₀(-0.01)° + ⁵C₁(-0.01)¹ + ⁵C₂ (-0.01)² + ………………..
= 1 – 5(0.01) + 10(0.0001) + …………
= 1 – 0.05 + 0.001
= 0.951

Question 2.
Expand (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ using Binomial theorem,
Solution:
Here (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ = [(1 + \(\frac{x}{2}\)) – \(\frac{2}{x}\)]⁴
Using Binomial theorem, the expression (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ can be expressed as

Again applying Binomial theorem, we have

Now putting the value of (1 + \(\frac{x}{2}\))⁴ and (1 + \(\frac{x}{3}\))³ in (1) we have

Question 3.
Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.
Solution:
Here, (3x² – 2ax + 3a²)³ = ((3x² – 2ax) + 3a²)³
Using Binomial theorem, the expression [(3x² – 2ax) + 3a²)³ can be expressed as
³C0(3x² – 2ax)³ + ³Cj(3x² – 2ax)² (3a²) + ³C2(3x² – 2ax)¹ (3a²)² + ³C3(3a²)³ = (3x² – 2ax)³ + 3(9x⁴ + 4a²x² – 12ax³)(3a²) + 3(3x² – 2ax) (9a⁴) + (27a⁶)
= (3x² – 2ax)³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶

Again applying Binomial theorem, we have
(3x² – 2ax)³ = ³C₀(3x²)³ + ³C₁(3x²)² (-2ax)¹ + ³C₂(3x²) (-2ax)² + ³C₃(-2ax)³
= 27x⁶ – 3 × 18ax⁵ + 3 × 12a²x⁴ – 8a³x³
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³

Now putting the value of (3x² – 2ax)³, we have
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Question 4.
Prove that C₀ + 3.C₁ + 5.C₂ + … + (2n + 1). C_n = (2n + 2). 2^n-1
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + … + (2n + 1).C_n ……….. (1)
By writing the terms in (1) in the reverse order we get
S = (2n + 1). C_n + (2n – 1) . C_n-1 + (2n – 3).C_n-2 + …….. + 3.C₁ + 1.C₀
= (2n + 1).C₀ + (2n – 1).C₁ + (2n – 3).C₂ + ……… + 3.C_n-1 + 1.C_n …………. (2)
On adding (1) and (2), we get
2S = (2n + 2).C₀ + (2n + 2).C₁ + … + (2n + 2).C_n-1 + (2n + 2) . C_n
= (2n + 2) (C₀ + C₁ + C₂ + … + C_n) = (2n + 2).2n Therefore, S = (2n + 2) 2^n-1

Question 5.
Prove that, for any real numbers a, d,
a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n = (2a + nd) . 2^n-1
Solution:
Let S = a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n ….. (1)
∴ S = (a + nd) . C_n + [a + (n – 1) d] . C_n-1 + … aC₀
⇒ S = (a + nd) . C₀ + [a + (n – 1)d]. + [a + (n – 2)d].C₂ + … + a . C_n ……….. (2)
(1) + (2) => 2S = (2a + nd) C₀ + (2a + nd) C₁ + (2a + nd) C₂ + … + (2a + nd) C_n
= (2a + nd) (C₀ + C₁ + C₂ + … + C_n)
= (2a + nd) 2n ⇒ S = (2a + nd) 2_n – 1.

Question 6.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Solution:
Let S = C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Now multiplying on both sides with (n + 1), we get
(n + 1) S = (n + 1)ⁿC₀ + \(\frac{1}{2}\) (n + 1)ⁿC₁ + —(n + 1)ⁿC₂ + … + ⁿC_n
= (n +1) + \(\frac{(\mathrm{n}+1) \mathrm{n}}{1 \cdot 2}+\frac{(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)}{1 \cdot 2 \cdot 3}\) + … + 1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺¹⁾C₃ + ……… + ⁽ⁿ⁺¹⁾C_n+1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺²⁾C₃ ……… + ⁽ⁿ⁺¹⁾C_n+1 – ⁽ⁿ⁺¹⁾C₀
= 2ⁿ⁺¹ -1
⇒ S = \(\frac{2^{n+1}-1}{n+1}\)
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)

Question 7.
If n is a positive integer and x is any non-zero real number, then prove that
C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}=\frac{(1+x)^{n+1}-1}{(n+1) x}\)
Solution:
Let S ≡ C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}\)

Multiplying both sides with be ‘x’

Question 8.
Prove that 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n + 2) C_n = (3n + 4)2^n-1
Solution:
Let S = 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n – 1) C_n-1 + (3n + 2) C_n ……(1)
S = (3n + 2) C₀ + (3n – 1)C₁ + (3n – 4) C₂ + + 5C_n-1 + 2.C_n … (2)
(∵ C_n = C₀.C_n-1 = Ci…)
Adding (1) 86 (2) we get 2S = (3n + 4)C0 + (3n + 4)Cj + (3n + 4)C2 + … + (3n + 4) Cn = (3n + 4) (C0 + Cj + C2 + … + Cn) = (3n + 4)2n
S = (3n + 4) . 2^n-1

Question 9.
Prove tnat C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ
Solution:
We know that,
(1 + x)ⁿ = C₀ + C₁x + C₂x² + C₃x² + ………… + C_nxⁿ
Put x = 2 in equation (1) we get,
(1 + 2)ⁿ = C₀ + C₁(2) + C₂(2)² + C₃(2)³ + ………. + C_n(2)ⁿ
∴ C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ

Question 10.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ
Put x = 1 ⇒ a₀ + a₁ + a₂ + ………… + a_2n = (1 + 1 + 1)ⁿ = 3ⁿ …….(1)
Put x = -1 ⇒ a₀ – a₁ + a₂ + ………. + a_2n = (1 – 1 + 1)ⁿ = 1 ………. (2)

a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ

ii) a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)
Solution:
(1) + (2) ⇒ 2(a₀ + a₂ + a₄ + ………… + a_2n) = 3ⁿ + 1
⇒ ∴ a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)

iii) a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)
Solution:
(1) – (2) ⇒ 2(a₁ + a₃ + a₅ + ….. + a_2n-1) = 3ⁿ – 1
⇒ ∴ a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)

iv) a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3^n-1
Solution:
Put x = 1
⇒ a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3ⁿ …(a)
Put x = ω
a₀ + a₁ω + a₂ω² + a₃ω³ + ………… + a_2nω²ⁿ = 0
Hint: 1 + ω + ω² = 0; ω³ = 1 ……………. (b)
Put x = ω²
⇒ a₀ + a₁ω² + a₂ω⁴ + a₃ω⁶ + ………… + a_2nω⁴ⁿ = 0 …..(c)
Adding (a), (b), (c)
∴ 3₀ + a₁(1 + ω + ω²) + a₂(1 + ω³ + ω⁶) + ……… + a_2n(1 + ω²ⁿ + ω⁴ⁿ) = 3ⁿ
⇒ 3a₀ + a₁(0) + a₂(0) + 3a₃ + ……. = 3ⁿ
∴ a₀ + a₃ + a₆ + ……… = \(\frac{3^{\mathrm{n}}}{3}\) = 3^n-1

Question 11.
If P and Q Me the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ
Solution:
We know that (x + a)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1a + ⁿC₂ x^n-2a² + ………………
(x + a)ⁿ = T₁ + T₂ + T₃ + T₄ + T₅ + T₆ + …………….
Given that P = (T₁ + T₃ + T₅ + T₇ + …) and Q = (T₂ + T₄ + T₆ + ………….)
⇒ (x + a)ⁿ = P + Q …………(1) and (x – a)ⁿ = P – Q ………….(2)

i) P² – Q² = (P + Q) (P – Q) = (x + a)ⁿ (x – a)ⁿ = [(x + a) (x – a)]ⁿ = (x² – a²)ⁿ

ii) 4PQ = (P + Q)² – (P – Q)²
= [(x + a)ⁿ]² – [(x – a)ⁿ]²
= (x + a)²ⁿ – (x – a)²ⁿ

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰, then prove that
i) a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + … + a₂₀ = 4¹⁰
Solution:
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰ ………….. (1)

i) Put x = 1 in eq. (1)
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + … + a₂₀
∴ a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰

ii) Put x = -1 in eq (1)
(1 – 3 – 2)¹⁰ = a₀ – a₁ + a₂ – a₃ + … + a₂₀
∴ a₀ – a₁ + a₂ – a₃ + … + a₂₀ = (-4)¹⁰ = 4¹⁰

Question 13.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{n!}\) 2ⁿ . xⁿ, where n is a positive integer.
Solution:
As 2n is even, the middle term of the expansion (1 + x)²ⁿ is i.e., (n + 1)^th term
Now T_n+1 = ²ⁿC_n (1)^2n-n (x)ⁿ = ²ⁿC_n xⁿ = \(\frac{(2 \mathrm{n})!}{\mathrm{n}!\mathrm{n}!}\)xⁿ

Question 14.
The second, third and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720, 1080 respectively. Find the value of x, a and n.
Solution:
Given expression is (a + x)ⁿ. General term T_r+1 = ⁿC_r a^n-r x^r
T₂ = T₁₊₁ = ⁿC₁ a^n-1 x¹ = 240 … (1)
T₃ = T₂₊₁ = ⁿC₂ a^n-2 x² = 720 … (2)
T₄ = T₃₊₁ = ⁿC₃ a^n-3 x³ = 1080 … (3)

⇒ 4n – 8 = 3n – 3
⇒ 4n – 3n = -3 + 8
∴ n = 5

Substitute n = 5 in (4) ⇒ 4x = 6a ⇒ x = 3a/2
Substitute x, n values in (1)
⇒ 5 a⁴\(\left(\frac{3 \mathrm{a}}{2}\right)\) = 240
⇒ a⁵ = \(\frac{240 \times 2}{15}\) = 32
⇒ a⁵ = 2⁵
⇒ a = 2

Substitute the values of n = 5 and a = 2 in (6)
⇒ x = \(\frac{3(2)}{2}\) = 3
∴ a = 2, x = 3, n = 5