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Class 11 NCERT Solutions

Chapter 8: Sequence and Series

Master the growth of progressions, the summation of infinite series, and the logic of numerical patterns with our step-by-step logic.

Exercise 8.1

1. Write the first five terms of the sequence whose $n$th term is $a_n = n(n+2)$.

We are given that $a_n = n(n+2)$. To find the first five terms, we substitute $n = 1, 2, 3, 4, 5$ one by one:

$ \begin{aligned} & a_1 = 1(1+2) = 3 \\ & a_2 = 2(2+2) = 8 \\ & a_3 = 3(3+2) = 15 \end{aligned} $

Substituting $n = 4$: $\quad a_4 = 4(4+2) = 24$

Substituting $n = 5$: $\quad a_5 = 5(5+2) = 35$

Therefore, the first five terms of the sequence are 3, 8, 15, 24, 35.

2. Write the first five terms of the sequence whose $n$th term is $\boldsymbol{a_n = \dfrac{n}{n+1}}$.

We are given $a_n = \dfrac{n}{n+1}$. Substituting $n = 1, 2, 3, 4, 5$ gives us:

$ \begin{array}{ll} a_1=\frac{1}{1+1}=\frac{1}{2}, & a_2=\frac{2}{2+1}=\frac{2}{3}, \quad a_3=\frac{3}{3+1}=\frac{3}{4} \\ a_4=\frac{4}{4+1}=\frac{4}{5}, \quad a_5=\frac{5}{5+1}=\frac{5}{6} \end{array} $

Therefore, the first five terms of the sequence are $\dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dfrac{4}{5},\ \dfrac{5}{6}$.

3. Write the first five terms of the sequence whose $n$th term is $a_n = 2^n$.

We are given $a_n = 2^n$. Plugging in $n = 1, 2, 3, 4, 5$:

$ \begin{array}{ll} a_1=2^1=2, & a_2=2^2=4, \quad a_3=2^3=8 \\ a_4=2^4=16, & a_5=2^5=32 \end{array} $

Therefore, the first five terms of the sequence are 2, 4, 8, 16, 32.

4. Write the first five terms of the sequence whose $n$th term is $a_n = \dfrac{2n-3}{6}$.

We are given $a_n = \dfrac{2n-3}{6}$. Substituting $n = 1, 2, 3, 4, 5$:

$ \begin{array}{ll} a_1=\frac{2 \times 1-3}{6}=-\frac{1}{6}, & a_2=\frac{2 \times 2-3}{6}=\frac{1}{6} \\ a_3=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}, & a_4=\frac{2 \times 4-3}{6}=\frac{5}{6} \\ a_5=\frac{2 \times 5-3}{6}=\frac{7}{6} & \end{array} $

Therefore, the first five terms are $-\dfrac{1}{6},\ \dfrac{1}{6},\ \dfrac{1}{2},\ \dfrac{5}{6},\ \dfrac{7}{6}$.

5. $\boldsymbol{a}_{\boldsymbol{n}}=(-\mathbf{1})^{\boldsymbol{n}-\mathbf{1}} \mathbf{5}^{\boldsymbol{n}+\mathbf{1}}$

Sol. Here $a_n=(-1)^{n-1} \cdot 5^{n+1}$ Putting $n=1,2,3,4$ and 5 , we have $$ \begin{aligned} & a_1=(-1)^0 \cdot 5^2=25, \quad a_2=(-1)^1 \cdot 5^3=-125 \\ & a_3=(-1)^2 \cdot 5^4=625, \quad a_4=(-1)^3 \cdot 5^5=-3125 \\ & a_5=(-1)^4 \cdot 5^6=15625 \end{aligned} $$Therefore, the required terms are 25, – 125, 625, – 3125, 15625.

6. $a_n=n \frac{n^2+5}{4}$.

Sol. Here $\quad a_n=\frac{n\left(n^2+5\right)}{4}$ Putting $n=1,2,3,4$ and 5 , we have $$ \begin{aligned} & a_1=\frac{1\left(1^2+5\right)}{4}=\frac{6}{4}=\frac{3}{2}, \quad a_2=\frac{2\left(2^2+5\right)}{4}=\frac{2 \times 9}{4}=\frac{9}{2}, \\ & a_3=\frac{3\left(3^2+5\right)}{4}=\frac{3 \times 14}{4}=\frac{21}{2}, \\ & a_4=\frac{4\left(4^2+5\right)}{4}=\frac{4 \times 21}{4}=21, \\ & a_5=\frac{5\left(5^2+5\right)}{4}=\frac{5 \times 30}{4}=\frac{75}{2} \end{aligned} $$Therefore, the required terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2}$.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n$th terms are:
7. $\boldsymbol{a}_{\boldsymbol{n}}=\mathbf{4 n}-\mathbf{3} ; \boldsymbol{a}_{\mathbf{1 7}}, \boldsymbol{a}_{\mathbf{2 4}}$

Sol. Here $\quad a_n=4 n-3$ Putting $n=17$, we have $a_{17}=4 \times 17-3=65$ Putting $n=24$, we have $a_{24}=4 \times 24-3=93$.

8. $a_n=\frac{n^2}{2^n} ; a_7$

Sol. Here $a_n=\frac{n^2}{2^n}$ Putting $n=7$, we have $a_7=\frac{7^2}{2^7}=\frac{49}{128}$.

9. Find the 9th term of the sequence $\boldsymbol{a_n = (-1)^{n-1}\, n^3}$.

We are given $a_n = (-1)^{n-1} \cdot n^3$. To find the 9th term, substitute $n = 9$:

$a_9 = (-1)^{8} \cdot 9^3 = (+1) \times 729 = 729.$

10. Find $a_{20}$ for the sequence $a_n = \dfrac{n(n-2)}{n+3}$.

We are given $a_n = \dfrac{n(n-2)}{n+3}$. Substituting $n = 20$:

$a_{20} = \frac{20 \times 18}{23} = \frac{360}{23}.$

11. Write the first five terms and the corresponding series for: $a_1 = 3,\; a_n = 3a_{n-1}+2$ for all $n > 1$.

We are given $a_1 = 3$ and the recurrence $a_n = 3a_{n-1} + 2$ for $n > 1$. We build each term step by step:

$n=2$: $a_2 = 3a_1 + 2 = 3 \times 3 + 2 = 11$

$n=3$: $a_3 = 3a_2 + 2 = 3 \times 11 + 2 = 35$

$n=4$: $a_4 = 3a_3 + 2 = 3 \times 35 + 2 = 107$

$n=5$: $a_5 = 3a_4 + 2 = 3 \times 107 + 2 = 323$

Therefore, the first five terms are 3, 11, 35, 107, 323.

∴ The corresponding series is $3 + 11 + 35 + 107 + 323 + \ldots$

12. Write the first five terms and the corresponding series for: $a_1 = -1,\; a_n = \dfrac{a_{n-1}}{n},\; n \geq 2$.

We are given $a_1 = -1$ and $a_n = \dfrac{a_{n-1}}{n}$. Computing each term:

$n=2$: $a_2 = \dfrac{a_1}{2} = \dfrac{-1}{2}$

$n=3$: $a_3 = \dfrac{a_2}{3} = \dfrac{-1/2}{3} = -\dfrac{1}{6}$

$n=4$: $a_4 = \dfrac{a_3}{4} = \dfrac{-1/6}{4} = -\dfrac{1}{24}$

$n=5$: $a_5 = \dfrac{a_4}{5} = \dfrac{-1/24}{5} = -\dfrac{1}{120}$

Therefore, the first five terms are $-1,\ -\dfrac{1}{2},\ -\dfrac{1}{6},\ -\dfrac{1}{24},\ -\dfrac{1}{120}$.

∴ The corresponding series is: $-1 + \left(\frac{-1}{2}\right) + \left(\frac{-1}{6}\right) + \left(\frac{-1}{24}\right) + \left(\frac{-1}{120}\right) + \ldots$

13. Write the first five terms and the corresponding series for: $a_1 = a_2 = 2,\; a_n = a_{n-1} – 1,\; n > 2$.

We are given $a_1 = a_2 = 2$ and $a_n = a_{n-1} – 1$ for $n > 2$. Step by step:

$n=3$: $a_3 = a_2 – 1 = 2 – 1 = 1$

$n=4$: $a_4 = a_3 – 1 = 1 – 1 = 0$

$n=5$: $a_5 = a_4 – 1 = 0 – 1 = -1$

Therefore, the first five terms are 2, 2, 1, 0, −1.

The corresponding series is $2 + 2 + 1 + 0 + (-1) + \ldots$

14. The Fibonacci sequence is defined by $1 = a_1 = a_2$ and $a_n = a_{n-1} + a_{n-2},\; n > 2$. Find $\dfrac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$.

We start with $a_1 = 1,\ a_2 = 1$ and use the recurrence $a_n = a_{n-1} + a_{n-2}$ to build the sequence:

$n=3$: $a_3 = a_2 + a_1 = 1 + 1 = 2$

$n=4$: $a_4 = a_3 + a_2 = 2 + 1 = 3$

$n=5$: $a_5 = a_4 + a_3 = 3 + 2 = 5$

$n=6$: $a_6 = a_5 + a_4 = 5 + 3 = 8$

Now we compute the required ratios:

For $n=1$: $\dfrac{a_2}{a_1} = \dfrac{1}{1} = 1$

For $n=2$: $\dfrac{a_3}{a_2} = \dfrac{2}{1} = 2$

For $n=3$: $\dfrac{a_4}{a_3} = \dfrac{3}{2}$

For $n=4$: $\dfrac{a_5}{a_4} = \dfrac{5}{3}$

For $n=5$: $\dfrac{a_6}{a_5} = \dfrac{8}{5}$

Exercise 8.2

1. Find the $20^{\text{th}}$ and $n^{\text{th}}$ terms of the G.P. $\dfrac{5}{2},\ \dfrac{5}{4},\ \dfrac{5}{8},\ \ldots$

We identify the first term and common ratio: $a = \dfrac{5}{2}$, $r = \dfrac{5/4}{5/2} = \dfrac{1}{2}$.

Using the formula $a_n = ar^{n-1}$ for a G.P.:

$ \begin{aligned} a_{20} &= ar^{19} = \frac{5}{2}\left(\frac{1}{2}\right)^{19} = \frac{5}{2} \times \frac{1}{2^{19}} = \frac{5}{2^{20}} \\ a_n &= ar^{n-1} = \frac{5}{2}\left(\frac{1}{2}\right)^{n-1} = \frac{5}{2} \times \frac{1}{2^{n-1}} = \frac{5}{2^n} \end{aligned} $

2. Find the $12^{\text{th}}$ term of a G.P. whose $8^{\text{th}}$ term is 192 and the common ratio is 2.

We are given $r = 2$. Let $a$ be the first term.

Since $a_8 = 192$, we have $ar^7 = 192$:

$ \begin{aligned} & a \times 2^7 = 192 \quad \therefore \quad a = \frac{192}{128} = \frac{3}{2} \\ & \therefore \quad a_{12} = ar^{11} = \frac{3}{2} \times 2^{11} = 3 \times 2^{10} = 3 \times 1024 = 3072 \end{aligned} $

3. The $5^{\text{th}}$, $8^{\text{th}}$ and $11^{\text{th}}$ terms of a G.P. are $p$, $q$ and $s$ respectively. Show that $q^2 = ps$.

Let $a$ be the first term and $r$ the common ratio. We express each given term:

$ \begin{array}{ll} a_5 = p \Rightarrow ar^4 = p \\ a_8 = q \Rightarrow ar^7 = q \\ a_{11} = s \Rightarrow ar^{10} = s \end{array} $

Now we compare both sides:

L.H.S.: $q^2 = (ar^7)^2 = a^2r^{14}$

R.H.S.: $ps = ar^4 \cdot ar^{10} = a^2r^{14}$

∴ L.H.S. = R.H.S., i.e., $q^2 = ps$. (Proved)

4. The $4^{\text{th}}$ term of a G.P. is the square of its second term, and the first term is $-3$. Determine its $7^{\text{th}}$ term.

Here $a = -3$. Let $r$ be the common ratio. We are given $a_4 = (a_2)^2$:

$ar^3 = (ar)^2 \Rightarrow ar^3 = a^2r^2$

Dividing both sides by $ar^2$:

$ \begin{aligned} r &= a = -3 \\ \therefore \quad a_7 &= ar^6 = (-3) \times (-3)^6 = -3 \times 729 = -2187 \end{aligned} $

5. Which term of the following sequences: (a) $2, 2\sqrt{2}, 4, \ldots$ is 128? (b) $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is 729? (c) $\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots$ is $\dfrac{1}{19683}$?

(a) We verify this is a G.P. with $a = 2$, $r = \sqrt{2}$ (since each ratio equals $\sqrt{2}$).

Let $a_n = 128$, so $ar^{n-1} = 128$:

$ \begin{aligned} 2 \times (\sqrt{2})^{n-1} &= 128 \\ (\sqrt{2})^{n-1} = 64 &= 2^6 = (\sqrt{2})^{12} \\ n – 1 &= 12 \quad \Rightarrow \quad n = 13 \end{aligned} $

∴ 128 is the 13th term.

(b) This is a G.P. with $a = \sqrt{3}$, $r = \sqrt{3}$.

Let $a_n = 729$:

$\sqrt{3} \times (\sqrt{3})^{n-1} = 729 \Rightarrow (\sqrt{3})^n = 3^6 = (\sqrt{3})^{12} \Rightarrow n = 12$

∴ 729 is the 12th term.

(c) This is a G.P. with $a = \dfrac{1}{3}$, $r = \dfrac{1}{3}$.

Let $a_n = \dfrac{1}{19683}$:

$\frac{1}{3}\left(\frac{1}{3}\right)^{n-1} = \frac{1}{19683} \Rightarrow \left(\frac{1}{3}\right)^n = \left(\frac{1}{3}\right)^9 \Rightarrow n = 9$

∴ $\dfrac{1}{19683}$ is the 9th term.

6. For what values of $x$ are the numbers $-\dfrac{2}{7},\ x,\ -\dfrac{7}{2}$ in G.P.?

For three numbers to be in G.P., the ratio of consecutive terms must be equal, i.e., $\dfrac{\text{II}}{\text{I}} = \dfrac{\text{III}}{\text{II}}$:

$ \begin{aligned} \frac{x}{-\frac{2}{7}} &= \frac{-\frac{7}{2}}{x} \end{aligned} $

Cross-multiplying:

$x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1 \quad \therefore \quad x = \pm 1$

7. Find the sum of the G.P.: $0.15,\ 0.015,\ 0.0015,\ \ldots$ up to 20 terms.

Here $a = 0.15 = \dfrac{15}{100}$, $r = \dfrac{0.015}{0.15} = \dfrac{1}{10} < 1$, $n = 20$.

Using $S_n = \dfrac{a(1-r^n)}{1-r}$:

$ \begin{aligned} S_{20} &= \frac{\frac{15}{100}\left[1-\left(\frac{1}{10}\right)^{20}\right]}{1 – \frac{1}{10}} = \frac{\frac{15}{100}\left[1-(0.1)^{20}\right]}{\frac{9}{10}} \\ &= \frac{10}{9} \times \frac{3}{20}\left[1-(0.1)^{20}\right] = \frac{1}{6}\left[1-(0.1)^{20}\right] \end{aligned} $

8. Find the sum of the G.P.: $\sqrt{7},\ \sqrt{21},\ 3\sqrt{7},\ \ldots$ up to $n$ terms.

Here $a = \sqrt{7}$, $r = \dfrac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} > 1$.

Using $S_n = \dfrac{a(r^n – 1)}{r-1}$:

$S_n = \frac{\sqrt{7}\left[(\sqrt{3})^n – 1\right]}{\sqrt{3}-1}$

Rationalising the denominator by multiplying numerator and denominator by $(\sqrt{3}+1)$:

$ \begin{aligned} &= \frac{\sqrt{7}(\sqrt{3}+1)\left[(\sqrt{3})^n-1\right]}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ &= \frac{\sqrt{7}(\sqrt{3}+1)\left[3^{n/2}-1\right]}{3-1} \\ &= \frac{\sqrt{7}}{2}(\sqrt{3}+1)\left(3^{n/2}-1\right) \end{aligned} $

9. Find the sum of the G.P.: $1,\ -a,\ a^2,\ -a^3,\ \ldots$ up to $n$ terms (if $a \neq -1$).

Here the first term $a_1 = 1$ and the common ratio $r = -a$.

$ S_n = \frac{a_1(1-r^n)}{1-r} = \frac{1\left[1-(-a)^n\right]}{1-(-a)} = \frac{1-(-a)^n}{1+a} $

Note: Since the sign of $r = -a$ is unknown, you may also use $S_n = \dfrac{a(r^n-1)}{r-1}$ — both forms are valid here.

10. Find the sum of the G.P.: $x^3,\ x^5,\ x^7,\ \ldots$ up to $n$ terms (if $x \neq \pm 1$).

Here $a = x^3$, $r = \dfrac{x^5}{x^3} = x^2$.

$S_n = \frac{a(1-r^n)}{1-r} = \frac{x^3\left[1-(x^2)^n\right]}{1-x^2} = \frac{x^3(1-x^{2n})}{1-x^2}$

11. Evaluate $\displaystyle\sum_{k=1}^{11}(2+3^k)$.

We split the sum into two parts:

$ \sum_{k=1}^{11}(2+3^k) = (2+3+3^2+\cdots+3^{11}) \text{ rearranged as:} $ $ = (2+2+\ldots\,11\text{ times}) + (3+3^2+3^3+\ldots\text{ to }11\text{ terms}) $

The first part gives $11 \times 2 = 22$. The second part is a G.P. with $a=3, r=3, n=11$:

$ = 22 + \frac{3(3^{11}-1)}{3-1} = 22 + \frac{3}{2}(3^{11}-1) $

12. The sum of the first three terms of a G.P. is $\dfrac{39}{10}$ and their product is 1. Find the common ratio and the terms.

Let the three terms in G.P. be $\dfrac{a}{r},\ a,\ ar$.

Their product equals 1:

$\frac{a}{r} \cdot a \cdot ar = 1 \Rightarrow a^3 = 1 \therefore a = 1$

Their sum equals $\dfrac{39}{10}$:

$\frac{1}{r} + 1 + r = \frac{39}{10} \Rightarrow \frac{1 + r + r^2}{r} = \frac{39}{10}$

Cross-multiplying and simplifying:

$10r^2 – 29r + 10 = 0 \Rightarrow (2r-5)(5r-2) = 0 \Rightarrow r = \frac{5}{2} \text{ or } \frac{2}{5}$

When $a=1,\ r = \dfrac{5}{2}$: the three terms are $\dfrac{2}{5},\ 1,\ \dfrac{5}{2}$.

When $a=1,\ r = \dfrac{2}{5}$: the three terms are $\dfrac{5}{2},\ 1,\ \dfrac{2}{5}$.

13. How many terms of the G.P. $3,\ 3^2,\ 3^3,\ \ldots$ are needed to give the sum 120?

Here $a = 3$, $r = 3 > 1$. Let $S_n = 120$:

$\frac{3(3^n – 1)}{3-1} = 120 \Rightarrow \frac{3}{2}(3^n – 1) = 120$ $3(3^n – 1) = 240 \Rightarrow 3^n – 1 = 80 \Rightarrow 3^n = 81 = 3^4 \Rightarrow n = 4$

∴ The sum of 4 terms of the G.P. equals 120.

14. The sum of the 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 $S_n$.

Let $a$ and $r$ be the first term and common ratio respectively.

From the given conditions:

$a + ar + ar^2 = 16 \Rightarrow a(1+r+r^2) = 16 \quad \cdots(i)$ $ar^3 + ar^4 + ar^5 = 128 \Rightarrow ar^3(1+r+r^2) = 128 \quad \cdots(ii)$

Dividing (ii) by (i) to eliminate $a$:

$r^3 = 8 = 2^3 \therefore r = 2$

Substituting $r = 2$ in (i):

$a(1+2+4) = 16 \Rightarrow 7a = 16 \therefore a = \frac{16}{7}$ $S_n = \frac{a(r^n-1)}{r-1} = \frac{\frac{16}{7}(2^n-1)}{1} = \frac{16}{7}(2^n-1)$

15. Given a G.P. with $a = 729$ and $7^{\text{th}}$ term 64, determine $S_7$.

Here $a = 729$. From $a_7 = 64$:

$ar^6 = 64 \Rightarrow 729\, r^6 = 64 \Rightarrow r^6 = \frac{64}{729} = \left(\pm\frac{2}{3}\right)^6 \therefore r = \pm\frac{2}{3}$

When $r = \dfrac{2}{3}$:

$S_7 = \frac{a(1-r^7)}{1-r} = \frac{729\left(1-\frac{128}{2187}\right)}{\frac{1}{3}} = 3 \times 729 \times \frac{2059}{2187} = 2059$

When $r = -\dfrac{2}{3}$:

$S_7 = \frac{729\left[1-\left(-\frac{2}{3}\right)^7\right]}{1-\left(-\frac{2}{3}\right)} = \frac{3}{5} \times 729\left[1+\frac{128}{2187}\right] = \frac{2187}{5} \times \frac{2315}{2187} = 463$

16. Find a G.P. for which the sum of the first two terms is $-4$ and the fifth term is 4 times the third term.

Let $a$ be the first term and $r$ the common ratio.

From $a_1 + a_2 = -4$: $\quad a(1+r) = -4 \quad\cdots(i)$

From $a_5 = 4a_3$: $\quad ar^4 = 4ar^2 \Rightarrow r^2 = 4 \therefore r = \pm 2$

When $r = 2$: From (i), $3a = -4 \Rightarrow a = -\dfrac{4}{3}$. G.P.: $-\dfrac{4}{3},\ -\dfrac{8}{3},\ -\dfrac{16}{3},\ \ldots$

When $r = -2$: From (i), $-a = -4 \Rightarrow a = 4$. G.P.: $4,\ -8,\ 16,\ -32,\ 64,\ \ldots$

17. If the $4^{\text{th}}$, $10^{\text{th}}$, and $16^{\text{th}}$ terms of a G.P. are $x$, $y$, $z$ respectively, show that $x$, $y$, $z$ are in G.P.

Let $a$ be the first term and $r$ the common ratio.

$ \begin{array}{ll} a_4 = x \Rightarrow ar^3 = x \\ a_{10} = y \Rightarrow ar^9 = y \\ a_{16} = z \Rightarrow ar^{15} = z \end{array} $

Computing the ratios:

$\frac{y}{x} = \frac{ar^9}{ar^3} = r^6 \quad \text{and} \quad \frac{z}{y} = \frac{ar^{15}}{ar^9} = r^6$

Since $\dfrac{y}{x} = \dfrac{z}{y}$ (each equals $r^6$), $x$, $y$, $z$ are in G.P. (Proved)

18. Find the sum to $n$ terms of the sequence $8,\ 88,\ 888,\ 8888,\ \ldots$

This is neither an A.P. nor a G.P., but we can relate it to a G.P. by factoring:

$8 + 88 + 888 + \ldots \text{ to } n \text{ terms} = 8(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})$

Multiplying and dividing by 9 to create all 9’s:

$= \frac{8}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})$ $= \frac{8}{9}\left[(10-1)+(10^2-1)+(10^3-1)+\ldots \text{ to } n \text{ terms}\right]$ $= \frac{8}{9}\left[\left(10+10^2+10^3+\ldots \text{ to }n\text{ terms}\right) – n\right]$

Using $S_n$ of G.P. with $\alpha = 10$, $r = 10 > 1$:

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

19. Find the sum of the products of corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \dfrac{1}{2}$.

We multiply the corresponding terms first:

$ (2)(128) + (4)(32) + (8)(8) + (16)(2) + (32)\left(\frac{1}{2}\right) = 256 + 128 + 64 + 32 + 16 $

This is a G.P. with $a = 256$, $r = \dfrac{1}{2} < 1$, $n = 5$:

$= \frac{256\left[1-\left(\frac{1}{2}\right)^5\right]}{1-\frac{1}{2}} = 2 \times 256 \times \frac{31}{32} = 496$

20. Show that the products of the corresponding terms of $a, ar, ar^2, \ldots, ar^{n-1}$ and $A, AR, AR^2, \ldots, AR^{n-1}$ form a G.P. Find the common ratio.

The products of corresponding terms are:

$(a)(A),\ (ar)(AR),\ (ar^2)(AR^2),\ \ldots,\ (ar^{n-1})(AR^{n-1})$

which simplifies to:

$aA,\ aA(rR),\ aA(rR)^2,\ \ldots,\ aA(rR)^{n-1}$

This is a G.P. with first term $aA$ and common ratio $rR$, since:

$\frac{\text{II}}{\text{I}} = \frac{aA(rR)}{aA} = rR \quad \text{and} \quad \frac{\text{III}}{\text{II}} = rR$

∴ The products form a G.P. with common ratio $\boldsymbol{rR}$. (Proved)

21. Find four numbers forming a G.P. in which the third term is greater than the first by 9, and the second term is greater than the fourth by 18.

Let the four numbers in G.P. be $a, ar, ar^2, ar^3$.

From the first condition ($T_3 – T_1 = 9$):

$ar^2 – a = 9 \Rightarrow a(r^2 – 1) = 9 \quad\cdots(i)$

From the second condition ($T_2 – T_4 = 18$):

$ar – ar^3 = 18 \Rightarrow ar(1 – r^2) = 18 \Rightarrow ar(r^2-1) = -18 \quad\cdots(ii)$

Dividing (ii) by (i) to eliminate $a$: $r = -2$.

Substituting in (i): $a(4-1) = 9 \Rightarrow a = 3$.

∴ The four numbers are 3, −6, 12, −24.

22. If the $p^{\text{th}}$, $q^{\text{th}}$ and $r^{\text{th}}$ terms of a G.P. are $a$, $b$, $c$ respectively, prove that $a^{q-r}\,b^{r-p}\,c^{p-q}=1$.

Let A be the first term and R the common ratio of the G.P.:

$\mathrm{AR}^{p-1}=a, \quad \mathrm{AR}^{q-1}=b, \quad \mathrm{AR}^{r-1}=c$

Substituting these into the L.H.S.:

$ \begin{aligned} a^{q-r}b^{r-p}c^{p-q} &= (\mathrm{AR}^{p-1})^{q-r} \cdot (\mathrm{AR}^{q-1})^{r-p} \cdot (\mathrm{AR}^{r-1})^{p-q} \\ &= \mathrm{A}^{q-r+r-p+p-q} \cdot \mathrm{R}^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \\ &= \mathrm{A}^0 \cdot \mathrm{R}^0 = 1 \times 1 = 1 \end{aligned} $

(The exponents of A and R both simplify to 0.) (Proved)

23. If the first and $n^{\text{th}}$ terms of a G.P. are $a$ and $b$ respectively, and $P$ is the product of $n$ terms, prove that $P^2 = (ab)^n$.

Let $r$ be the common ratio. Writing the product $P$ in forward order:

$\mathrm{P} = a(ar)\left(ar^2\right)\ldots\left(\frac{b}{r^2}\right)\left(\frac{b}{r}\right)b \quad\cdots(i)$

Now writing the same product in reverse order:

$\mathrm{P} = b\left(\frac{b}{r}\right)\left(\frac{b}{r^2}\right)\ldots(ar^2)(ar)a \quad\cdots(ii)$

Multiplying (i) and (ii): each paired product gives $ab$, and there are $n$ such pairs:

$\mathrm{P}^2 = (ab)(ab)(ab)\ldots(ab) = (ab)^n$

(Proved)

24. Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text{th}}$ to $(2n)^{\text{th}}$ term is $\dfrac{1}{r^n}$.

Let $a$ and $r$ be the first term and common ratio of the G.P.

$S_1 = \text{Sum of first }n\text{ terms} = \frac{a(1-r^n)}{1-r} \quad\cdots(i)$ $S_2 = \text{Sum from }(n+1)^{\text{th}}\text{ to }(2n)^{\text{th}}\text{ term} = ar^n + ar^{n+1} + \ldots + ar^{2n-1}$

This is a G.P. with first term $ar^n$, ratio $r$, and $n$ terms:

$S_2 = \frac{ar^n(1-r^n)}{1-r} \quad\cdots(ii)$

Dividing (i) by (ii):

$\frac{S_1}{S_2} = \frac{a(1-r^n)}{1-r} \times \frac{1-r}{ar^n(1-r^n)} = \frac{1}{r^n}$

(Proved)

25. If $a, b, c, d$ are in G.P., show that $(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2$.

Let $r$ be the common ratio: $b = ar,\ c = ar^2,\ d = ar^3$.

L.H.S.:

$ \begin{aligned} &= (a^2+a^2r^2+a^2r^4)(a^2r^2+a^2r^4+a^2r^6) \\ &= a^2(1+r^2+r^4) \cdot a^2r^2(1+r^2+r^4) \\ &= a^4r^2(1+r^2+r^4)^2 \end{aligned} $

R.H.S.:

$ \begin{aligned} (ab+bc+cd)^2 &= (a^2r+a^2r^3+a^2r^5)^2 \\ &= \left[a^2r(1+r^2+r^4)\right]^2 = a^4r^2(1+r^2+r^4)^2 \end{aligned} $

L.H.S. = R.H.S. (Proved)

26. Insert two numbers between 3 and 81 so that the resulting sequence is a G.P.

Let the two inserted numbers be $G_1$ and $G_2$ so that $3, G_1, G_2, 81$ form a G.P. of four terms.

Let $r$ be the common ratio. Using $a_4 = ar^3$:

$3r^3 = 81 \Rightarrow r^3 = 27 = 3^3 \therefore r = 3$ $G_1 = T_2 = ar = 3 \times 3 = 9, \quad G_2 = T_3 = ar^2 = 3 \times 9 = 27$

The two numbers to be inserted are 9 and 27.

27. Find $n$ such that $\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the geometric mean between $a$ and $b$.

The G.M. between $a$ and $b$ is $\sqrt{ab}$. Setting up the equation:

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab} = a^{1/2}b^{1/2}$

Cross-multiplying:

$a^{n+1}+b^{n+1} = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2}$

Rearranging and factoring:

$a^{n+1/2}(a^{1/2}-b^{1/2}) = b^{n+1/2}(a^{1/2}-b^{1/2})$

Dividing both sides by $(a^{1/2}-b^{1/2})$:

$a^{n+1/2} = b^{n+1/2} \Rightarrow \left(\frac{a}{b}\right)^{n+1/2} = 1 = \left(\frac{a}{b}\right)^0$ $\therefore n + \frac{1}{2} = 0 \quad \Rightarrow \quad n = -\frac{1}{2}$

28. The sum of two numbers is 6 times their geometric mean. Show that the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.

Let the two numbers be $a$ and $b$, $a > b$. We are given $a + b = 6\sqrt{ab}$, which gives:

$\frac{a+b}{2\sqrt{ab}} = \frac{3}{1}$

Applying componendo and dividendo:

$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{4}{2} \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 2$

Taking square roots:

$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{2}$

Applying componendo and dividendo again:

$\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$

Squaring both sides:

$\frac{a}{b} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}} \quad \Rightarrow \quad a:b = (3+2\sqrt{2}):(3-2\sqrt{2})$

(Proved)

29. If A and G are the A.M. and G.M. respectively between two positive numbers, prove that the numbers are $A \pm \sqrt{(A+G)(A-G)}$.

Let the two quantities be $a$ and $b$, with $a > b$.

$\mathrm{A} = \frac{a+b}{2} \Rightarrow a+b = 2\mathrm{A} \quad\cdots(i)$ $\mathrm{G} = \sqrt{ab} \Rightarrow ab = \mathrm{G}^2 \quad\cdots(ii)$

Using the identity $(a-b)^2 = (a+b)^2 – 4ab$ and substituting from (i) and (ii):

$(a-b)^2 = 4\mathrm{A}^2 – 4\mathrm{G}^2 \Rightarrow a-b = 2\sqrt{\mathrm{A}^2-\mathrm{G}^2} \quad\cdots(iii)$

Adding (i) and (iii) and dividing by 2:

$a = \mathrm{A} + \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$

Subtracting (iii) from (i) and dividing by 2:

$b = \mathrm{A} – \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$

∴ The two numbers are $\mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$. (Proved)

30. Bacteria in a certain culture double every hour. If there were 30 bacteria originally, how many will be present at the end of the $2^{\text{nd}}$, $4^{\text{th}}$, and $n^{\text{th}}$ hour?

At $t = 0$, bacteria count is 30. Since the count doubles each hour, the number at the end of the $1^{\text{st}}, 2^{\text{nd}}, 3^{\text{rd}}, \ldots$ hours forms a G.P.: $30 \times 2,\ 30 \times 2^2,\ 30 \times 2^3, \ldots$

(i) At the end of the $2^{\text{nd}}$ hour: $30 \times 2^2 = 120$

(ii) At the end of the $4^{\text{th}}$ hour: $30 \times 2^4 = 480$

(iii) At the end of the $n^{\text{th}}$ hour: $30 \times 2^n$

31. What will ₹500 amount to in 10 years at 10% per annum compound interest?

For compound interest, the amount grows by a fixed ratio each year. Amount at the end of each year:

End of year 1: $₹\,500(1.1)$

End of year 2: $₹\,500(1.1)^2$

End of year 3: $₹\,500(1.1)^3$, and so on.

These form a G.P. Therefore, the required amount at the end of 10 years is:

$₹\,500(1.1)^{10}$

32. If the A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, find the quadratic equation.

Let the roots be $\alpha$ and $\beta$.

$\text{A.M.} = 8 \Rightarrow \frac{\alpha+\beta}{2} = 8 \therefore \alpha+\beta = 16$ $\text{G.M.} = 5 \Rightarrow \sqrt{\alpha\beta} = 5 \therefore \alpha\beta = 25$

The required quadratic equation is $x^2 – Sx + P = 0$, i.e.:

$x^2 – (\alpha+\beta)x + \alpha\beta = 0 \Rightarrow \boxed{x^2 – 16x + 25 = 0}$

Miscellaneous Exercise

1. Show that the sum of $(m+n)^{\text{th}}$ and $(m-n)^{\text{th}}$ terms of an A.P. equals twice the $m^{\text{th}}$ term.

Let $a$ be the first term and $d$ the common difference of the A.P.

$ \begin{aligned} a_{m+n} + a_{m-n} &= [a+(m+n-1)d] + [a+(m-n-1)d] \\ &= 2a + (m+n-1+m-n-1)d \\ &= 2a + (2m-2)d \\ &= 2[a+(m-1)d] = 2a_m \end{aligned} $

∴ The sum of the $(m+n)^{\text{th}}$ and $(m-n)^{\text{th}}$ terms equals twice the $m^{\text{th}}$ term. (Proved)

2. If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Let the three numbers in A.P. be $a-d,\ a,\ a+d$.

From the sum condition: $(a-d) + a + (a+d) = 24 \Rightarrow 3a = 24 \therefore a = 8$

From the product condition:

$(8-d) \times 8 \times (8+d) = 440 \Rightarrow 64 – d^2 = 55 \Rightarrow d^2 = 9 \therefore d = \pm 3$

When $a = 8, d = 3$: numbers are $5, 8, 11$.

When $a = 8, d = -3$: numbers are $11, 8, 5$.

∴ The numbers are 5, 8, 11 or 11, 8, 5.

3. Let $S_1$, $S_2$, $S_3$ be the sums of $n$, $2n$, $3n$ terms of an A.P. respectively. Show that $S_3 = 3(S_2 – S_1)$.

Let $a$ be the first term and $d$ the common difference.

$S_1 = \frac{n}{2}[2a+(n-1)d], \quad S_2 = \frac{2n}{2}[2a+(2n-1)d], \quad S_3 = \frac{3n}{2}[2a+(3n-1)d]$

Computing R.H.S.:

$ \begin{aligned} 3(S_2 – S_1) &= 3\left[\frac{2n}{2}\{2a+(2n-1)d\} – \frac{n}{2}\{2a+(n-1)d\}\right] \\ &= \frac{3n}{2}[4a+(4n-2)d – 2a-(n-1)d] \\ &= \frac{3n}{2}[2a+(3n-1)d] = S_3 \end{aligned} $

∴ $S_3 = 3(S_2 – S_1)$. (Proved)

4. Find the sum of all numbers between 200 and 400 which are divisible by 7.

The smallest multiple of 7 greater than 200 is $203$; the largest multiple of 7 less than 400 is $399$.

Required sum $= 203 + 210 + 217 + \ldots + 399$. This is an A.P. with $a = 203$, $d = 7$, $l = 399$.

Finding $n$: $203 + (n-1) \times 7 = 399 \Rightarrow n – 1 = 28 \Rightarrow n = 29$.

$\text{Sum} = \frac{n}{2}(a+l) = \frac{29}{2}(203+399) = \frac{29}{2} \times 602 = 29 \times 301 = 8729$

5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Let $S_1$ = sum of integers divisible by 2, $S_2$ = sum divisible by 5, $S_3$ = sum divisible by both (i.e. by 10).

$S_1 = 2+4+\ldots+100 = \frac{50}{2}(2+100) = 2550$ $S_2 = 5+10+\ldots+100 = \frac{20}{2}(5+100) = 1050$ $S_3 = 10+20+\ldots+100 = \frac{10}{2}(10+100) = 550$

By inclusion-exclusion:

$\text{Required sum} = S_1 + S_2 – S_3 = 2550 + 1050 – 550 = 3050$

6. Find the sum of all two-digit numbers which, when divided by 4, yield remainder 1.

Two-digit multiples of 4 start from 12. Adding 1 to each gives: $13, 17, 21, \ldots, 97$.

This is an A.P. with $a = 13$, $d = 4$, $l = 97$.

Finding $n$: $13 + (n-1) \times 4 = 97 \Rightarrow n – 1 = 21 \Rightarrow n = 22$.

$\text{Sum} = \frac{22}{2}(13+97) = 11 \times 110 = 1210$

7. If $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{N}$, $f(1) = 3$, and $\displaystyle\sum_{x=1}^n f(x) = 120$, find $n$.

Using the functional equation with $f(1) = 3$:

$f(2) = f(1)f(1) = 9$; $f(3) = f(1)f(2) = 27$; and so on.

So $f(1) + f(2) + f(3) + \ldots + f(n) = 3 + 9 + 27 + \ldots = 120$.

This is a G.P. with $a = 3$, $r = 3$:

$\frac{3(3^n – 1)}{3-1} = 120 \Rightarrow 3(3^n – 1) = 240 \Rightarrow 3^n = 81 = 3^4 \therefore n = 4$

8. The sum of some terms of a G.P. is 315, with first term 5 and common ratio 2. Find the last term and the number of terms.

Here $a = 5$, $r = 2 > 1$. Setting $S_n = 315$:

$\frac{5(2^n-1)}{2-1} = 315 \Rightarrow 5(2^n-1) = 315 \Rightarrow 2^n – 1 = 63 \Rightarrow 2^n = 64 = 2^6 \therefore n = 6$

Last term $= a_6 = ar^5 = 5 \times 2^5 = 5 \times 32 = \mathbf{160}$.

9. The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio.

Let $r$ be the C.R. with $a = 1$. Then $a_3 = r^2$ and $a_5 = r^4$:

$r^2 + r^4 = 90 \Rightarrow r^4 + r^2 – 90 = 0$

Substituting $t = r^2$:

$t^2 + t – 90 = 0 \quad \text{Discriminant: } D = 1+360 = 361$ $t = \frac{-1 \pm 19}{2} = 9 \text{ or } -10$

Since $r^2 = -10$ gives imaginary values, we take $r^2 = 9$:

$\therefore r = \pm 3$

10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in order, we get an A.P. Find the numbers.

Let the three G.P. numbers be $a, ar, ar^2$.

$a(1+r+r^2) = 56 \quad\cdots(i)$

After subtracting: $a-1,\ ar-7,\ ar^2-21$ are in A.P., so $\text{II} – \text{I} = \text{III} – \text{II}$:

$a(1-2r+r^2) = 8 \Rightarrow a(r-1)^2 = 8 \quad\cdots(ii)$

Dividing (i) by (ii):

$\frac{1+r+r^2}{1-2r+r^2} = 7 \Rightarrow 7(r-1)^2 = 1+r+r^2 \Rightarrow 6r^2-15r+6=0$ $\Rightarrow 2r^2-5r+2=0 \Rightarrow (r-2)(2r-1)=0 \therefore r=2 \text{ or } \frac{1}{2}$

When $r = 2$: $a = 8$, numbers are 8, 16, 32.

When $r = \dfrac{1}{2}$: $a = 32$, numbers are 32, 16, 8.

11. A G.P. consists of an even number of terms. If the sum of all terms is 5 times the sum of the odd-placed terms, find the common ratio.

Let the G.P. have $2n$ terms with first term $a$ and common ratio $r$.

$\frac{a(1-r^{2n})}{1-r} = 5 \times \frac{a(1-r^{2n})}{1-r^2}$

Simplifying (dividing both sides by $\dfrac{a(1-r^{2n})}{1-r}$):

$1 = \frac{5}{1+r} \Rightarrow 1+r = 5 \therefore r = 4$

12. The sum of the first four terms of an A.P. is 56 and the sum of the last four terms is 112. If the first term is 11, find the number of terms.

Here $a = 11$. Using $S_4 = 56$:

$\frac{4}{2}[2(11)+(4-1)d] = 56 \Rightarrow 2(22+3d) = 56 \Rightarrow 3d = 6 \therefore d = 2$

Let $l$ be the last term. The last four terms are $l-3d, l-2d, l-d, l$:

$4l – 6d = 112 \Rightarrow 4l – 12 = 112 \Rightarrow 4l = 124 \therefore l = 31$

Using $a_n = l$: $11 + (n-1) \times 2 = 31 \Rightarrow n – 1 = 10 \therefore n = 11$.

∴ The number of terms is 11.

13. If $\dfrac{a+bx}{a-bx} = \dfrac{b+cx}{b-cx} = \dfrac{c+dx}{c-dx}\ (x \neq 0)$, show that $a, b, c, d$ are in G.P.

From the first two equal members, cross-multiplying:

$(a+bx)(b-cx) = (a-bx)(b+cx)$

Expanding and simplifying:

$2b^2x = 2acx \Rightarrow b^2 = ac \Rightarrow \frac{b}{a} = \frac{c}{b} \quad\cdots(i)$

Similarly, from the second and third members:

$c^2 = bd \Rightarrow \frac{c}{b} = \frac{d}{c} \quad\cdots(ii)$

From (i) and (ii): $\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$, so $a, b, c, d$ are in G.P. (Proved)

14. Let $S$, $P$, $R$ be the sum, product, and sum of reciprocals of $n$ terms of a G.P. Prove that $P^2 R^n = S^n$.

Let $a$ and $r$ be the first term and common ratio.

$S = \frac{a(1-r^n)}{1-r} \quad\cdots(i)$ $P = a^n \cdot r^{1+2+\ldots+(n-1)} = a^n \cdot r^{\frac{n(n-1)}{2}} \quad\cdots(ii)$

For $R$ (a G.P. with first term $\frac{1}{a}$ and ratio $\frac{1}{r}$):

$R = \frac{1-r^n}{ar^{n-1}(1-r)} \quad\cdots(iii)$

Computing L.H.S. $= P^2 R^n$:

$= \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 \left(\frac{1-r^n}{ar^{n-1}(1-r)}\right)^n = \frac{a^{2n}r^{n(n-1)}(1-r^n)^n}{a^n r^{n(n-1)}(1-r)^n} = \frac{a^n(1-r^n)^n}{(1-r)^n}$

R.H.S. $= S^n = \left(\dfrac{a(1-r^n)}{1-r}\right)^n = \dfrac{a^n(1-r^n)^n}{(1-r)^n}$

∴ L.H.S. = R.H.S. (Proved)

15. The $p^{\text{th}}$, $q^{\text{th}}$, $r^{\text{th}}$ terms of an A.P. are $a$, $b$, $c$ respectively. Show that $(q-r)a + (r-p)b + (p-q)c = 0$.

Let A be the first term and $d$ the common difference.

$\mathrm{A}+(p-1)d=a, \quad \mathrm{A}+(q-1)d=b, \quad \mathrm{A}+(r-1)d=c$

Substituting into the L.H.S.:

$ \begin{aligned} &= (q-r)(A+pd-d) + (r-p)(A+qd-d) + (p-q)(A+rd-d) \\ &= 0 \quad (\text{all terms cancel in pairs}) \end{aligned} $

(Proved)

16. If $a\!\left(\tfrac{1}{b}+\tfrac{1}{c}\right)$, $b\!\left(\tfrac{1}{c}+\tfrac{1}{a}\right)$, $c\!\left(\tfrac{1}{a}+\tfrac{1}{b}\right)$ are in A.P., prove that $a, b, c$ are in A.P.

The given terms are in A.P., which means:

$\frac{a(b+c)}{bc},\ \frac{b(c+a)}{ca},\ \frac{c(a+b)}{ab} \text{ are in A.P.}$

Adding 1 to each term (A.P. property is preserved):

$\frac{ab+ac+bc}{bc},\ \frac{bc+ab+ca}{ca},\ \frac{ca+bc+ab}{ab} \text{ are in A.P.}$

Dividing each term by $ab+bc+ca$:

$\frac{1}{bc},\ \frac{1}{ca},\ \frac{1}{ab} \text{ are in A.P.}$

Multiplying each term by $abc$:

$a,\ b,\ c \text{ are in A.P.} \quad \textbf{(Proved)}$

17. If $a, b, c, d$ are in G.P., prove that $(a^n+b^n)$, $(b^n+c^n)$, $(c^n+d^n)$ are in G.P.

Let $r$ be the common ratio: $b = ar,\ c = ar^2,\ d = ar^3$.

For the three expressions to be in G.P., we need $\dfrac{b^n+c^n}{a^n+b^n} = \dfrac{c^n+d^n}{b^n+c^n}$.

L.H.S.:

$\frac{(ar)^n+(ar^2)^n}{a^n+(ar)^n} = \frac{a^nr^n(1+r^n)}{a^n(1+r^n)} = r^n$

R.H.S.:

$\frac{(ar^2)^n+(ar^3)^n}{(ar)^n+(ar^2)^n} = \frac{a^nr^{2n}(1+r^n)}{a^nr^n(1+r^n)} = r^n$

Since L.H.S. = R.H.S. = $r^n$, the three expressions are in G.P. (Proved)

18. If $a, b$ are roots of $x^2 – 3x + p = 0$ and $c, d$ are roots of $x^2 – 12x + q = 0$, where $a, b, c, d$ form a G.P., prove that $(q+p):(q-p) = 17:15$.

From the first equation: $a+b = 3$ and $ab = p$.

From the second equation: $c+d = 12$ and $cd = q$.

Let $r$ be the common ratio of the G.P.: $b = ar,\ c = ar^2,\ d = ar^3$.

From $a(1+r) = 3$ and $ar^2(1+r) = 12$, dividing gives $r^2 = 4$.

Now computing the ratio:

$\frac{q+p}{q-p} = \frac{cd+ab}{cd-ab} = \frac{a^2r^5+a^2r}{a^2r^5-a^2r} = \frac{r^4+1}{r^4-1} = \frac{16+1}{16-1} = \frac{17}{15}$

∴ $(q+p):(q-p) = 17:15$. (Proved)

19. The ratio of A.M. to G.M. of two positive numbers is $m:n$. Show that $a:b = \left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$.

Given $\dfrac{a+b}{2\sqrt{ab}} = \dfrac{m}{n}$. Applying componendo and dividendo:

$\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \frac{m+n}{m-n} \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{m+n}}{\sqrt{m-n}}$

Applying componendo and dividendo again:

$\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

Squaring both sides:

$\frac{a}{b} = \frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}} = \frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$

(Proved)

20. If $a, b, c$ are in A.P.; $b, c, d$ are in G.P.; and $\dfrac{1}{c}, \dfrac{1}{d}, \dfrac{1}{e}$ are in A.P., prove that $a, c, e$ are in G.P.

From the three given conditions:

A.P.: $2b = a + c \Rightarrow b = \dfrac{a+c}{2} \quad\cdots(i)$

G.P.: $c^2 = bd \quad\cdots(ii)$

A.P.: $\dfrac{2}{d} = \dfrac{1}{c} + \dfrac{1}{e} \Rightarrow d = \dfrac{2ce}{c+e} \quad\cdots(iii)$

Substituting (i) and (iii) into (ii) to eliminate $b$ and $d$:

$c^2 = \frac{a+c}{2} \cdot \frac{2ce}{c+e} = \frac{(a+c)ce}{c+e}$

Cross-multiplying and dividing by $c$:

$c(c+e) = (a+c)e \Rightarrow c^2 + ce = ae + ce \Rightarrow c^2 = ae$

Since $c \cdot c = a \cdot e$, we have $\dfrac{c}{a} = \dfrac{e}{c}$, confirming $a, c, e$ are in G.P. (Proved)

21. Find the sum up to $n$ terms: (i) $5 + 55 + 555 + \ldots$ (ii) $0.6 + 0.66 + 0.666 + \ldots$

(i) Factoring out 5 and then multiplying/dividing by 9:

$S_n = 5(1+11+111+\ldots) = \frac{5}{9}(9+99+999+\ldots)$ $= \frac{5}{9}\left[(10+10^2+\ldots \text{ to }n\text{ terms}) – n\right] = \frac{50}{81}(10^n-1) – \frac{5n}{9}$

(ii) Factoring out 6 and multiplying/dividing by 9:

$S_n = \frac{6}{9}\left[(1+1+\ldots n\text{ times}) – \left(\frac{1}{10}+\frac{1}{100}+\ldots n\text{ terms}\right)\right]$ $= \frac{2}{3}\left[n – \frac{1}{9}\left(1-\frac{1}{10^n}\right)\right] = \frac{2}{27}\left[9n – 1 + 10^{-n}\right]$

22. Find the $20^{\text{th}}$ term of the series $2\times4 + 4\times6 + 6\times8 + \ldots$ to $n$ terms.

Each term is the product of the $k^{\text{th}}$ terms of two A.P.s: $2, 4, 6, \ldots$ and $4, 6, 8, \ldots$

The $20^{\text{th}}$ term of $2,4,6,\ldots$ is $2 + 19\times2 = 40$.

The $20^{\text{th}}$ term of $4,6,8,\ldots$ is $4 + 19\times2 = 42$.

$a_{20} = 40 \times 42 = 1680$

23. Find the sum of the first $n$ terms of the series: $3 + 7 + 13 + 21 + 31 + \ldots$

Write $S_n$ and shift one place, then subtract to find $a_n$:

$S_n = 3+7+13+21+\ldots+a_n$ $S_n = \quad 3+7+13+\ldots+a_{n-1}+a_n$

On subtracting:

$0 = 3 + (4+6+8+\ldots \text{ to }(n-1)\text{ terms}) – a_n$ $a_n = 3 + \frac{n-1}{2}[8+2(n-2)] = 3 + (n-1)(n+2) = n^2+n+1$

Therefore:

$S_n = \sum n^2 + \sum n + n = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n = \frac{n}{3}(n^2+3n+5)$

24. If $S_1$, $S_2$, $S_3$ are the sums of first $n$ natural numbers, their squares, and their cubes respectively, show that $9S_2^2 = S_3(1+8S_1)$.

We use the standard formulas:

$S_1 = \frac{n(n+1)}{2}, \quad S_2 = \frac{n(n+1)(2n+1)}{6}, \quad S_3 = \frac{n^2(n+1)^2}{4}$

L.H.S.:

$9S_2^2 = 9\cdot\frac{n^2(n+1)^2(2n+1)^2}{36} = \frac{n^2(n+1)^2(2n+1)^2}{4}$

R.H.S.:

$S_3(1+8S_1) = \frac{n^2(n+1)^2}{4}\left(1+4n(n+1)\right) = \frac{n^2(n+1)^2}{4}(4n^2+4n+1) = \frac{n^2(n+1)^2(2n+1)^2}{4}$

L.H.S. = R.H.S. (Proved)

25. Find the sum up to $n$ terms: $\dfrac{1^3}{1} + \dfrac{1^3+2^3}{1+3} + \dfrac{1^3+2^3+3^3}{1+3+5} + \ldots$

The $n^{\text{th}}$ term of the series is:

$a_n = \frac{1^3+2^3+\ldots+n^3}{1+3+5+\ldots \text{ to }n\text{ terms}} = \frac{\left\{\frac{n(n+1)}{2}\right\}^2}{\frac{n}{2}(2+2n-2)} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$

Therefore:

$S_n = \frac{1}{4}\sum(n+1)^2 = \frac{1}{4}\sum(n^2+2n+1) = \frac{1}{4}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\cdot\frac{n(n+1)}{2}+\frac{n}{4}$ $= \frac{n}{24}(2n^2+9n+13)$

26. Show that $\dfrac{1\times2^2+2\times3^2+\ldots+n\times(n+1)^2}{1^2\times2+2^2\times3+\ldots+n^2\times(n+1)} = \dfrac{3n+5}{3n+1}$.

Numerator: The $n^{\text{th}}$ term is $n(n+1)^2 = n^3+2n^2+n$:

$\text{Numerator} = \sum n^3 + 2\sum n^2 + \sum n = \frac{n(n+1)(n+2)(3n+5)}{12}$

Denominator: The $n^{\text{th}}$ term is $n^2(n+1) = n^3+n^2$:

$\text{Denominator} = \sum n^3 + \sum n^2 = \frac{n(n+1)(n+2)(3n+1)}{12}$

Dividing:

$\frac{\text{Numerator}}{\text{Denominator}} = \frac{3n+5}{3n+1} \quad \textbf{(Proved)}$

27. A farmer buys a tractor for ₹12,000. He pays ₹6,000 cash and the balance in annual instalments of ₹500 plus 12% interest on the unpaid amount. How much does the tractor cost him in total?

Balance after cash payment $= ₹6,000$. Number of instalments $= \dfrac{6000}{500} = 12$.

Interest with each instalment decreases as the unpaid principal reduces by ₹500 each time:

Interest on 1st instalment (on ₹6000): $₹\dfrac{6000\times12}{100} = ₹720$

Interest on 2nd instalment (on ₹5500): $₹660$

Interest on 3rd instalment (on ₹5000): $₹600$, and so on.

Total interest $= 720 + 660 + 600 + \ldots$ (12 terms; A.P. with $a=720$, $d=-60$):

$= \frac{12}{2}[2(720) + 11(-60)] = 6(1440-660) = 6 \times 780 = ₹4680$

Total cost of tractor $= ₹12{,}000 + ₹4{,}680 = \mathbf{₹16{,}680}$.

28. Shamshad Ali buys a scooter for ₹22,000. He pays ₹4,000 cash and the balance in annual instalments of ₹1,000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Balance $= ₹18,000$. Number of instalments $= 18$.

1st instalment $= ₹1000 + \dfrac{18000\times10}{100} = ₹1000 + ₹1800 = ₹2800$

2nd instalment $= ₹1000 + ₹1700 = ₹2700$ (balance becomes ₹17,000)

3rd instalment $= ₹1000 + ₹1600 = ₹2600$, and so on.

Total paid in instalments (A.P. with $a=2800$, $d=-100$, $n=18$):

$= \frac{18}{2}[2(2800)+17(-100)] = 9(5600-1700) = 9 \times 3900 = ₹35{,}100$

Total cost $= ₹4{,}000 + ₹35{,}100 = \mathbf{₹39{,}100}$.

29. A person writes a letter to four friends, each of whom writes to four others, and so on. It costs 50 paise to mail one letter. Find the total postage spent when the 8th set of letters is mailed.

Total number of letters in 8 sets $= 4 + 16 + 64 + \ldots$ to 8 terms (G.P. with $a = r = 4$):

$= \frac{4(4^8-1)}{4-1} = \frac{4}{3}(65536-1) = \frac{4 \times 65535}{3} = 4 \times 21845 = 87380$

Total postage:

$= ₹\left(\frac{50}{100} \times 87380\right) = ₹43{,}690$

30. A man deposits ₹10,000 at 5% simple interest annually. Find the amount in the 15th year and the total amount after 20 years.

The amounts at the end of years 1, 2, 3, … are ₹10,500; ₹11,000; ₹11,500; … forming an A.P. with $a = 10500$, $d = 500$.

Amount in the 15th year = amount at the end of 14 years $= a_{14}$:

$a_{14} = 10500 + 13 \times 500 = ₹17{,}000$

Amount after 20 years $= a_{20}$:

$a_{20} = 10500 + 19 \times 500 = ₹20{,}000$

31. A machine costs ₹15,625 and depreciates by 20% each year. Find its estimated value at the end of 5 years.

Each year the machine retains $\left(1 – \dfrac{1}{5}\right) = \dfrac{4}{5}$ of its value.

After $n$ years, the value $= V_0 \left(\dfrac{4}{5}\right)^n$.

After 5 years:

$= 15625 \times \left(\frac{4}{5}\right)^5 = 15625 \times \frac{1024}{3125} = 5 \times 1024 = ₹5{,}120$

32. 150 workers were engaged to finish a job in $k$ days. 4 workers dropped out each day starting from the second day. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Total work $= 150k$ (worker-days). Let the actual number of days taken $= n = k + 8$.

The workers present on each day form an A.P.: $150, 146, 142, \ldots$ with $a=150$, $d=-4$.

Setting total actual work equal to planned work:

$\frac{n}{2}[300+(n-1)(-4)] = 150(n-8)$ $n(304-4n) = 300(n-8) \Rightarrow -4n^2+4n+2400=0 \Rightarrow n^2-n-600=0$ $\Rightarrow (n-25)(n+24) = 0$

Since $n$ cannot be negative, $n = \mathbf{25}$ days.