TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P :

Solution:

Question 2.
Find the
i) 30^th term of the A.P. : 10, 7, 4, …… (A.P. June 15)
ii) 11^th term of the A.P. : -3, \(\frac{-1}{2}\) ,2, …….
Solution:
i) Given A.P. = 10, 7, 4, ….
a₁ = 10; d = a₂ – a₁ = 7 – 10 = – 3
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1)(-3)
= 10 + 29 × (- 3)
= 10 – 87
= -77

ii) Given A.P. = -3, \(\frac{-1}{2}\), 2,……..
a₁ = -3; d = a₂ – a₁ = \(\frac{-1}{2}\) – (-3)
= \(\frac{-1}{2}\) + 3 = = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
a_n = a + (n – 1)d
a₁₁ = -3 + (11 – 1) × (\(\frac{5}{2}\))
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25 = 22

Question 3.
Find the respective terms for the following APs.

i) a₁ = 2; a₃ = 26, find a₂.
Solution:
Given : a₁ = a = 2 —— (1)
a₃ = a + 2d = 26 —— (2)
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a₂ = a + d = 2 + 12 = 14

ii) a₂ = 13; a₄ = 3, find a₁, a₃.
Solution:
Given : a₂ = a + d = 13 —— (1)
a₄ = a + 3d = 3 —- (2)
Solving equations (1) and (2);

Substituting, d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18
i.e., a₁ = 18
a₃ = a + 2d = 18 + 2(-5)
= 18 – 10 = 8

iii) a₁ = 5; a₄ = 9\(\frac{1}{2}\), a₂, a₃.
Solution:
Given : a₁ = a = 5 —– (1)
a₄ = a + 3d = 9\(\frac{1}{2}\) —– (2)
Solving equations (1) and (2);

iv) a₁ = – 4; a₆ = 6, find a₂, a₃, a₄, a₅.
Solution:
Given : a₁ = a = – 4 —– (1)
a₆ = a + 5d = 6 —– (2)
Solving equations (1) and (2);
(-4) + 5d = 6 ⇒ 5d = 6 + 4
⇒ 5d = 10 ⇒ d = \(\frac{10}{5}\) = 2
Now, a₂ = a + d = -4 + 2 = -2;
a₃ = a + 2d = -4 + 2 × 2
= -4 + 4 = 0;
a₄ = a + 3d = (-4) + 3 × 2
= -4 + 6 = 2;
a₅ = a + 4d = -4 + 4 × 2
= -4 + 8 = 4

v) a₂ = 38; a₆ = -22, find a₁, a₃, a₄, a₅.
Solution:
Given : a₂ = a + d = 38 —– (1)
a₆ = a + 5d = -22 —– (2)
Subtracting (2) from (1) we get

Now substituting, d = -15 in equation (1), we get
a + (-15) = 38 ⇒ a = 38 + 15 = 53 Thus,
a₁ = a = 53;
a₃ = a + 2d = 53 + 2 × (-15) = 53 – 30 = 23;
a₄ = a + 3d = 53 + 3 × (-15) = 53 – 45 = 8;
a₅ = a + 4d = 53 + 4 × (-15) = 53 – 60 = -7

Question 4.
Which term of the AP :
3, 8, 13, 18, ……. is 78 ?
Solution:
Given : 3, 8, 13, 18,
Here a = 3; d = a₂ – a₁ = 8 – 3 = 5
Let ’78’ be the nth term of the given A.P.
∴ a_n = a + (n – 1) d
78 = 3 + (n – 1)5
78 = 3 + 5n – 5
⇒ 5n = 78 + 2 ⇒ n = \(\frac{80}{5}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs :
i) 7, 13, 19, ……., 205 (A.P. Mar. ’16)
Solution:
Given : A.P : 7, 13, 19, ………
Here a₁ = a = 7; d = a₂ – a₁ = 13 – 7 = 6
Let 205 be the n^th term of the given A.P.
Then, a_n= a + (n – 1) d
205 = 7 + (n – 1) 6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, ………, -47.
Solution:
Given A.P = 18, 15\(\frac{1}{2}\), 13, ….
Here a₁ = a = 18
d = a₂ – a₁ = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the n^th term of the given A.P.
a_n = a + (n – 1) × d
⇒ -47 = \(\frac{18 \times 2+(n-1)(-5)}{2}\)
⇒ – 94 = 36 – 5n + 5
⇒ 5n = 94 + 41 ⇒ n = \(\frac{135}{5}\) = 27
i.e., 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP : 11, 8, 5, 2, ….
Solution:
Given A.P. = 11, 8, 5, 2,
Here a = a₂ – a₁ = 8 – 11 = -3
If possible, take -150 as the n,sup>th term of the given A.P.
a_n = a + (n – 1) d
-150 = 11 + (n – 1) × (- 3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = -150
⇒ 3n = 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31^st term of an A.P. whose 11^th term is 38 and the 16th term is 73.
Solution:
Given : An A.P. whose

Substituting d = 7 in the equation (1) we get,
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = – 32
Now, the 31^st term = a + 30d
= (- 32) + 30 × 7
= – 32 + 210 = 178

Question 8.
If the 3^rd and the 9^th terms of an A.P are 4 and – 8 respectively, which term of this A.P is zero ?
Solution:
Given : An A.P whose

Substituting d = – 2 in equation (1) we get a + 2 × (- 2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let n^th term of the given A.P be equal to zero.
a_n = a + (n – 1) d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5^th term of the given A.P is zero.

Question 9.
The 17^th term of an A.P exceeds its 10^th term by 7. Find the common difference.
Solution:
Given an A.P. in which a₁₇ = a₁₀ + 7
⇒ a₁₇ – a₁₀ = 7
We know that a_n = a + (n – 1) d

Question 10.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Solution:
Let the first A.P be : a, a + d, a + 2d,
Second A.P be : b, b + d, b + 2d, b + 3d,….
Also, general term, a_n = a + (n – 1) d
Given that, a₁₀₀ – b₁₀₀ = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000 terms,

∴ The difference between their 1000th terms is (a – b) = 100.

Note : If the common difference for any two A.Ps are equal then difference n^th terms of two A.Ps is same for all natural values of n.

Question 11.
How many three-digit numbers are divisible by 7 ?
Solution:
The list of three digit numbers divisible by 7 , is,
105, 112, 119, ………, 994
Here a = 105; d = 7; a_n = 994
a_n = a + (n – 1) d
994 = 105 + (n – 1) 7
(n – 1)7 = 889
n -1 = \(\frac{889}{7}\)
n – 1 = 127
∴ n = 127 + 1 = 128
There are 128 three digit numbers which are divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Given numbers : 10 to 250

∴ Multiples of 4 between 10 and 250 are
First term : 10 + (4 – 2) = 12
Last term : 250 – 2 = 248
∴ 12, 16, 20, 24, ………, 248
a = a₁ = 12; d = 4: a_n = 248
a_n = a + (n – 1) d
248 = 12 + (n – 1) × 4
⇒ (n – 1) 4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the n^th terms of two APs : 63, 65, 67, … and 3, 10, 17, …. equal ?
Solution:
Given : The first A.P is 63, 65, 67, ….
where a = 63, d = a₂ – a₁
⇒ d = 65 – 63 = 2
and the second A.P is 3, 10, 17, ….
where a = 3, d = a₂ – a₁ = 10 – 3 = 7
Suppose the n^th terms of the two A.Ps are equal, where a_n = a + (n – 1) d
⇒ 63 + (n -1) 2 = 3 + (n – 1) 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 +4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13^th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Solution:
Given : An A.P in which
a₃ = a + 2d = 16 —– (1)
and a₇ = a₅ + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A. P is a, a + d, a + 2d, a + 3d, …
⇒ 4, 4 + 6, 4 + 12, 4 + 18, …
⇒ A.P: 4, 10, 16, 22,….

Question 15.
Find the 20^th term from the end of the AP : 3, 8, 13,…….., 253.
Solution:
Given : An A.P : 3, 8, 13, …., 253
Here a = a₁ = 3
d = a₂ – a₁ = 8 – 3 = 5
a_n = 253, where 253 is the last term
a_n = a + (n – 1) d
∴ 253 = 3 + (n – 1) 5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20^th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a₃₂ = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155
= 158

Question 16.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the (A.P. Mar. ’15)
Solution:
Given an A.P in which
a₄ + a₈ = 24 ⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 —– (1)
and a₆ + a₁₀ = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 —– (2)

Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a , a + d, a + 2d, ……..
i.e., – 13, (- 13 + 5), (- 13 + 2 × 5)
⇒ -13, -8, -3, ………

Question 17.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach Rs. 7000?
Solution:
Given : Salary of Subba Rao in 1995 = ₹ 5000
Annual increment = ₹ 200
i.e., His salary increases by Rs. 200 every year.

Clearly 5000, 5200, 5400, ……… forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached ₹ 7000
after x – years,
i.e., a_n = 7000
But, a_n = a + (n – 1) d
7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ n – 1 = \(\frac{2000}{200}\)
= 10
⇒ n = 10 + 1 = 11
∴ In 11^th year his salary reached ₹ 7000.