AP Inter 1st Year Maths Exercise 8b Solutions

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
\(\frac{S_n}{S_{2 n}-S_n}=\frac{\frac{a\left(r^{11}-1\right)}{r-1}}{\frac{a\left(r^{2 n}-1\right)}{r-1}-\frac{a\left(r^n-1\right)}{r-1}}\)

\(\frac{r^n-1}{\left(r^{2 n}-1\right)-\left(r^n-1\right)}=\frac{r^n-1}{r^{2 n}-r^n}\) \(\frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}^{\mathrm{n}}\left(\mathrm{r}^{\mathrm{n}}-1\right)}=\frac{1}{\mathrm{r}^{\mathrm{n}}}\)

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.