ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.5

 Exercise 9.5

Question 1

Find the sum of:

(i) 20 terms of the series 2 + 6 + 18 + …

(ii) 10 terms of series 1 + √3 + 3 + …

(iii) 6 terms of the GP. $1,-\frac{2}{3}, \frac{4}{9}, \ldots$

(iv) 20 terms of the GP. 0.15, 0.015, 0.0015,…

(v) 100 terms of the series 0.7 + 0.07 + 0.007 +…

(vi) 5 terms and n terms of the series $1+\frac{2}{3}+\frac{4}{9}+\ldots$

(vii) n terms of the G.P. √7, √21, 3√7, …

(viii)n terms of the G.P. 1, – a, a², – a³, … (a ≠ – 1)

(ix) n terms of the G.P. $x^{3}, x^{5}, x^{7}$, … (x ≠ ±1).

Sol :

(i) 2 + 6 + 18 + … 20 terms

Here, a = 2, r = 3, n = 20, r > 1

$\mathrm{S}_{20}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{2\left(3^{20}-1\right)}{3-1}=\frac{2\left(3^{20}-1\right)}{2}=3^{20}-1$

(ii) $1+\sqrt{3}+3+\ldots 10$ terms

Here, $a=1, r=\sqrt{3}, n=10, r>1$

$\mathrm{S}_{10}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{1\left[(\sqrt{3})^{10}-1\right]}{\sqrt{3}-1}$

$=\frac{\left(\sqrt{3}^{10}-1\right)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

(Rationalising the denominator)

$=\frac{\left(3^{5}-1\right)(\sqrt{3}+1)}{3-1}=\frac{(243-1)(\sqrt{3}+1)}{2}$

$=\frac{242(\sqrt{3}+1)}{2}=121(\sqrt{3}+1)$

(iii) $1,-\frac{2}{3}, \frac{4}{9}, \ldots 6$ terms

Here, $a=1, r=\frac{-2}{3}, n=6, r<1$

$\mathrm{S}_{6}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{1\left[1-\left(\frac{-2}{3}\right)^{6}\right]}{1+\frac{2}{3}}$

$=\frac{3}{5}\left(1-\frac{(-2)^{6}}{3^{6}}\right)$

$=\frac{3}{5}\left[1-\frac{64}{729}\right]=\frac{3}{5}\left[\frac{729-64}{729}\right]$

$=\frac{3}{5} \times \frac{665}{729}=\frac{133}{243}$

(iv) 0.15,0.015,0.0015, ... upto 20 terms

Here, $a=0.15, r=\frac{1}{10}, n=20$

$\mathrm{S}_{20}=\frac{a\left(1-r^{n}\right)}{1-r}$ $(\because r<1)$

$=\frac{0.15\left[1-\left(\frac{1}{10}\right)^{20}\right]}{1-\frac{1}{10}}$

$=\frac{15 \times 10}{100 \times 9}\left[1-\frac{1}{10^{20}}\right]$

$=\frac{5}{30}\left[1-\frac{1}{10^{20}}\right]=\frac{1}{6}\left[1-\frac{1}{10^{20}}\right]$

(v) 0.7+0.07+0.007+... 100 terms

Here, $a=0.7, r=\frac{1}{10}, n=100, r<1$

$\mathrm{S}_{100}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{0.7\left[1-\left(\frac{1}{10}\right)^{100}\right]}{1-\frac{1}{10}}$

$=\frac{7 \times 10}{10 \times 9}\left(1-\frac{1}{10^{100}}\right)=\frac{7}{9}\left(1-\frac{1}{10^{100}}\right)$

(vi) $1+\frac{2}{3}+\frac{4}{9}+\ldots n$ terms, 5 terms

Here, $a=1, r=\frac{2}{3} \div 1=\frac{2}{3}, n=5, n,(r<1)$

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

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

and $\mathrm{S}_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=3\left[1-\frac{32}{243}\right]$

$=3\left[\frac{243-32}{243}\right]=\frac{211}{81}$

(vii) $\sqrt{7}, \sqrt{21}, 3 \sqrt{\overline{7}}, \ldots$

Here, $a=\sqrt{7}, r=\frac{\sqrt{21}}{\sqrt{7}}=\frac{\sqrt{7} \times \sqrt{3}}{\sqrt{7}}=\sqrt{3}$

$r>1, n=n$ terms

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \Rightarrow \mathrm{S}_{n}=\frac{\sqrt{7}\left[(\sqrt{3})^{n}-1\right]}{\sqrt{3}-1}$

Rationalising the denominator.

$\Rightarrow \mathrm{S}_{n}=\frac{\sqrt{7}\left[(\sqrt{3})^{n}-1\right]}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3+1}}$

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

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

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

(viii) $1,-a, a^{2},-a^{3}, \ldots(a \neq-1)$ upto $n$ terms

Here, a=1, r=-a

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

(ix) $x^{3}, x^{5}, x^{7}, \ldots(x \neq\pm 1)$

Here, $a=x^{3}, r=x^{2}$

$\therefore \mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{x^{3}\left[1-\left(x^{2}\right)^{n}\right]}{1-x^{2}}$ if $r<1$

$=\frac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}$

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

$=\frac{x^{3}\left(x^{2 n}-1\right)}{x^{2}-1}$

Question 2

Find the sum of the first 10 terms of the geometric series

√2 + √6 + √18 + ….

Sol :

√2 + √6 + √18 + ….

Here, a = √2 , r = √3, r > 1

$\therefore \mathrm{S}_{10}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{\sqrt{2}\left[(\sqrt{3})^{10}-1\right]}{\sqrt{3}-1}=\frac{\sqrt{2}}{\sqrt{3}-1}\left[(3)^{5}-1\right]$

$=\frac{\sqrt{2}}{\sqrt{3}-1}(243-1)=\frac{\sqrt{2}}{\sqrt{3}-1} \times 242$

$=\frac{\sqrt{2}(\sqrt{3}+1) \times 242}{(\sqrt{3}-1)(\sqrt{3}+1)}$

(Rationalising the denominator)

$=\frac{242(\sqrt{6}+\sqrt{2})}{3-1}=\frac{121(\sqrt{6}+\sqrt{2})}{2}$

$=121(\sqrt{6}+\sqrt{2})$

Question 3

Find the sum of the series $81-27+9 \ldots-\frac{1}{27}$

Sol :

$81-27+9 \ldots-\frac{1}{27}$

Here, $a=81, r=\frac{-27}{81}=\frac{-1}{3}, l=\frac{-1}{27}, r<1$

$S_{n}=\frac{a-l r}{1-r}$

$=\frac{81+\frac{1}{27} \times \frac{-1}{3}}{1+\frac{1}{3}}$

$=\frac{81-\frac{1}{81}}{\frac{4}{3}}$

$=\frac{6561-1}{81 \times \frac{4}{3}}$

$=\frac{6560 \times 3}{81 \times 4}=\frac{1640}{27}$

Question 4

The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.

Sol :

In a G.P.

$\mathrm{T}_{\mathrm{n}}=128$

$\mathrm{~S}_{\mathrm{n}}=255$

r=2

Let a be the first term, then

$\mathrm{T}_{n}=a r^{n-1}$

$\Rightarrow 128=a 2^{n-1}$

$\Rightarrow a=\frac{128}{2^{n-1}}$..(i)

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow 255=\frac{a\left(2^{n}-1\right)}{2-1}$

$\Rightarrow 255=a\left(2^{n}-1\right)$

$\Rightarrow a=\frac{255}{2^{n}-1}$..(ii)

From (i) and (ii)

$\frac{255}{2^{n}-1}=\frac{128}{2^{n-1}}$

$\Rightarrow 255 \times 2^{n-1}=128\left(2^{n}-1\right)$

$\Rightarrow 255 \times 2^{n-1}=128 \times 2^{n}-128$

$\frac{255 \times 2^{n}}{2}=128 \times 2^{n}-128$

$\Rightarrow 255 \times 2^{n}=256 \times 2^{n}-256$

$\Rightarrow 256 \times 2^{n}-255 \times 2^{n}=256$

$\Rightarrow 2^{n}=256=2^{8}$

Comparing, we get

n=8

Now, $128=a \cdot 2^{7}$

$\Rightarrow 128=a \times 128$

$\Rightarrow a=\frac{128}{128}=1$

$\therefore a=1$

Question 5

If the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P.

Sol :

Sum of first 6 terms of a G.P. = 9x The of first 3 terms

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

$\therefore \mathrm{S}_{6}=9 \times a_{3}$

$\Rightarrow \mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

$\mathrm{S}_{6}=\frac{a\left(r^{6}-1\right)}{r-1}$ and $\mathrm{S}_{3}=\frac{a\left(r^{3}-1\right)}{r-1}$

$\therefore \frac{a\left(r^{6}-1\right)}{r-1}=9 \times \frac{a\left(r^{3}-1\right)}{r-1}$

$\Rightarrow r^{6}-1=9\left(r^{3}-1\right)$

$\Rightarrow \frac{r^{6}-1}{r^{3}-1}=9$

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

$\Rightarrow r^{3}+1=9$

$\Rightarrow r^{3}=9-1=8=(2)^{2}$

$\therefore r=2$

$\therefore$ Common ratio $=2$

Question 6

A G.P. consists of an even number of terms. If the sum of all the terms is 3 times the sum of the odd terms, then find its common ratio.

Sol :

Let the G.P. be a, $a, a r, a r^{2}, \ldots a r^{2 n-1}$

These are 2n in number, which is an even number

A.T.Q.

$\Rightarrow a+a r+a r^{2}+\ldots+a r^{2 n-1}$

$=3\left(a+a r^{2}+a r^{4}+\ldots+a r^{2 n-2}\right)$

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

$\left[\because\right.$ Using $\left.\mathrm{S}^{n}=\frac{a-l r}{l-r}\right]$

$\Rightarrow 1-r^{2 n}=3\left[\frac{1-r^{2 n}}{1+r}\right]$

$\Rightarrow 1+r=3 \Rightarrow r=3-1 \Rightarrow r=2$

$\therefore$ Common ratio is 2

Question 7

(i) How many terms of the G.P. $3,3^{2}, 3^{3}, \ldots$ are needed to give the sum 120?

(ii) How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?

Solution:

In G.P.

(i) $3,3^{2}, 3^{3}$

Sum $=120,$ Here, $a=3, r=\frac{3^{2}}{3}=3, r>1$

Let number of terms in G.P. be n, then

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=120$

$\Rightarrow \frac{3\left(3^{n}-1\right)}{3-1}=120$

$\Rightarrow \frac{3\left(3^{n}-1\right)}{2}=120$

$\Rightarrow 3^{n}-1=\frac{120 \times 2}{3}=80$

$3^{n}=80+1=81=3^{4}$

$\therefore n=4$

$\therefore$ Number of terms =4

(ii) G.P. is 1,4,16,..

Sum=341

Here, $a=1, r=\frac{4}{1}=4, r>1$

Let number of terms be n, then

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=341$

$\Rightarrow \frac{1\left(4^{n}-1\right)}{4-1}=341$

$\Rightarrow \frac{1\left(4^{n}-1\right)}{4-1}=341$

$\Rightarrow \frac{4^{n}-1}{3}=341$

$\Rightarrow 4^{n}-1=341 \times 3=1023$

$4^{n}=1023+1=1024=2^{10}=4^{5}$

$\begin{array}{l|l}2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline & 1\end{array}$

$\therefore n=5$

$\therefore$ Number of terms $=5$

Question 8

How many terms of the GP. 1, √2 > 2, 2 √2 , … are required to give a sum of 1023( √2 + 1)?

Sol :

GP. 1, √2 > 2, 2 √2 , …

Sum = 1023 (√2 + 1)

Here, a = 1, r = √2 . r > 1

Let number of terms be n, then

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=1023(\sqrt{2}+1)$

$\Rightarrow \frac{1\left[(\sqrt{2})^{n}-1\right]}{\sqrt{2}-1}=1023(\sqrt{2}+1)$

$\Rightarrow(\sqrt{2})^{n}-1=1023(\sqrt{2}+1)(\sqrt{2}-1)$

$\Rightarrow(\sqrt{2})^{n}-1=1023\left[(\sqrt{2})^{2}-(1)^{2}\right]$

$\Rightarrow(\sqrt{2})^{n}-1=1023(2-1)=1023$

$(\sqrt{2})^{n}=1023+1=1024=2^{10}$ or $(\sqrt{2})^{20}$

$\begin{array}{l|l}2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline &1\end{array}$

Comparing, we get

n=20

Question 9

How many terms of the  $\frac{2}{9}-\frac{1}{3}+\frac{1}{2}+\ldots$  will make the sum $\frac{55}{72}$ ?

Sol :

G.P. is $\frac{2}{9}-\frac{1}{3}+\frac{1}{2}+\ldots$

sum $\frac{55}{72}$

Here, $a=\frac{2}{9}, r=\frac{-1}{3} \times \frac{9}{2}=\frac{-3}{2}, r<1$

Let number of terms be n

$\therefore \mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{55}{72}$

$\Rightarrow \frac{\frac{2}{9}\left[1-\left(\frac{-3}{2}\right)^{n}\right]}{1+\frac{3}{2}}=\frac{55}{72}$

$\Rightarrow 1-\left(\frac{-3}{2}\right)^{n}=\frac{55}{72} \times \frac{5}{2} \times \frac{9}{2}=\frac{275}{32}$

$\Rightarrow 1-(-1)^{n}\left(\frac{3}{2}\right)^{n}=\frac{275}{32}$

$\Rightarrow 1+1\left(\frac{3}{2}\right)^{n}=\frac{275}{32}$

$\Rightarrow\left(\frac{3}{2}\right)^{n}=\frac{275}{32}-1=\frac{275-32}{32}=\frac{243}{32}$

$=\left(\frac{3}{2}\right)^{5}$

Comparing, we get

n=5

$\therefore$ Number of terms $=5$

Question 10

The 2nd and 5th terms of a geometric series are $-\frac{1}{2} \text { and sum } \frac{1}{16}$  respectively. Find the sum of the series upto 8 terms.

Sol :

In a GP

$a_{2}=-\frac{1}{2} \text { and } a_{5}=\frac{1}{16}$

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

$\therefore a_{2}=a r^{n-1}=a r^{2-1}=a r=\frac{-1}{2}$...(i)

$a_{5}=a r^{5-1}=a r^{4}=\frac{1}{16}$..(ii)

Dividing (ii) by (i),

$r^{3}=\frac{1}{16} \div\left(\frac{-1}{2}\right)=\frac{1}{16} \times \frac{-2}{1}=\frac{-1}{8}$

$=\left(\frac{-1}{2}\right)^{3}$

$\therefore r=\frac{-1}{2}$

and $a r=\frac{-1}{2} \Rightarrow a \times\left(\frac{-1}{2}\right)=\frac{-1}{2}$

$\Rightarrow a=\frac{-1}{2} \times \frac{-2}{1}=1$

$a=1,$ and $r=\frac{-1}{2}$

$\mathrm{Now}, \mathrm{S}_{8}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{1\left[1-\left(\frac{-1}{2}\right)^{8}\right]}{1+\frac{1}{2}}$

$=\frac{1-\frac{1}{256}}{\frac{3}{2}}$

$=\frac{255}{256} \times \frac{2}{3}=\frac{510}{768}=\frac{85}{128}$

Question 11

The first term of a G.P. is 27 and 8th term is $\frac{1}{81}$ . Find the sum of its first 10 terms.

Sol :

In a G.P.

First term (a) = 27

$a_{8}=81$

Let r be the common ratio, then

$a r^{n-1}=a r^{8-1}=a r^{7}=\frac{1}{81}$

$\Rightarrow 27 r^{7}=\frac{1}{81}$

$\Rightarrow r^{7}=\frac{1}{81 \times 27}=\frac{1}{2187}=\frac{1}{(3)^{7}}$

$\therefore r=\frac{1}{3}$

Now, $S_{10}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{27\left[1-\left(\frac{1}{3}\right)^{10}\right]}{1-\frac{1}{3}}$

$=\frac{27\left[1-\frac{1}{3^{10}}\right]}{\frac{3-1}{3}}$

$=\frac{27 \times 3}{2}\left[1-\frac{1}{3^{10}}\right]=\frac{81}{2}\left[1-\frac{1}{3^{10}}\right]$

Question 12

Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728

Sol :

Common ratio of a G.P. = 3

and last term = 486

and sum of terms = 728

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{a\left(3^{n}-1\right)}{3-1}$

$=\frac{a\left(3^{n}-1\right)}{2}=728$

$a\left(3^{n}-1\right)=728 \times 2=1456$...(i)

l=486

$\Rightarrow a r^{n-1}=a \cdot 3^{n-1}=486$

$\Rightarrow a\left[\frac{3^{n}}{3}\right]=486$

$\Rightarrow a 3^{n}=486 \times 3=1458$..(ii0)

But $a\left(3^{n}-1\right)=1456$ [From (i)]

$a 3^{n}-a=1456$

$1458-a=1456$ [From (ii)]

$\therefore a=1458-1456=2$

Hence first term $=2$

Question 13

In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.

Sol :

In a GP.

First term (a) = 7, last term (l) = 448

and sum = 889

Let r be the common ratio, then

$l=a r^{n-1} \Rightarrow 7 r^{n-1}=448$

$r^{n-1}=\frac{448}{7}=64$ ..(i)

and sum $=\frac{a\left(r^{n}-1\right)}{r-1}=889$

$\Rightarrow \frac{7\left(r^{n}-1\right)}{r-1}=889$

$\Rightarrow \frac{r^{n}-1}{r-1}=\frac{889}{-7}=127$..(ii)

From (i)

$\frac{r^{n}}{r}=64 \Rightarrow r^{n}=64 r$

From (ii) $\frac{64 r-1}{r-1}=127$

$\Rightarrow 64 r-1=127 r-127$

$\Rightarrow 127 r-64 r=-1+127$

$\Rightarrow 63 r=126$

$\Rightarrow r=\frac{126}{63}=2$

Hence common ratio =2

Question 14

Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.

Sol :

In a G.P.

Common ratio = 3

$\mathrm{S}_{7}=2186=\frac{a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow 2186=\frac{a\left(3^{n}-1\right)}{3-1}$

$\Rightarrow 2186=\frac{a\left(3^{n}-1\right)}{2}$

$\Rightarrow 4372=a\left(3^{7}-1\right)$

$\Rightarrow 4372=a(2187-1)$

$\Rightarrow 4372=2186 a$

$\Rightarrow a=\frac{4372}{2186}=2$

$a_{3}=a r^{3-1}=a r^{2}=2 \times 3^{2}=2 \times 9=18$

Question 15

If the first term of a G.P. is 5 and the sum of first three terms is $\frac{31}{5}$, find the common ratio.

Sol :

In a G.P.

First term (a) = 5

$S_{3}=\frac{31}{5}=\frac{a\left(r^{3}-1\right)}{r-1}=\frac{5\left(r^{3}-1\right)}{r-1}$

$\frac{r^{3}-1}{r-1}=\frac{31}{5 \times 5}=\frac{31}{25}$

$\Rightarrow \frac{(r-1)\left(r^{2}+r+1\right)}{r-1}=\frac{31}{25}$

$r^{2}+r+1=\frac{31}{25}$

$\Rightarrow 25 r^{2}+25 r+25=31$

$\Rightarrow 25 r^{2}+25 r-6=0$

$\Rightarrow 25 r^{2}+30 r-5 r-6=0$

$\Rightarrow 5 r(5 r+6)-1(5 r+6)=0$

$\Rightarrow(5 r+6)(5 r-1)=0$

Either 5r+6=0, then $r=\frac{-6}{5}$

or 5r-1=0, then $r=\frac{1}{5}$

Hence common ratio $=\frac{1}{5}$ or $\frac{-6}{5}$

Question 16

The sum of first three terms of a GP. is to the sum of first six terms as 125 : 152. Find the common ratio of the GP.

Sol :

$S_{3} \div S_{6}=125: 152$

Let r be the common ratio and a be the first number, then

$\frac{a\left(r^{3}-1\right)}{r-1}: \frac{a\left(r^{6}-1\right)}{r-1}=125: 152$

$\left(r^{3}-1\right):\left(r^{6}-1\right)=125: 152$

$\left(r^{3}-1\right):\left(r^{3}+1\right)\left(r^{3}-1\right)=125: 152$

$1:\left(r^{3}+1\right)=125: 152$

$\left(r^{3}+1\right) \times 125=152 \times 1$

$125 r^{3}+125=152$

$125 r^{3}=152-125=27$

$r^{3}=\frac{27}{125}=\left(\frac{3}{5}\right)^{3}$

$\therefore r=\frac{3}{5}$

$\therefore$ Common ratio $=\frac{3}{5}$

Question 17

Find the sum of the products of the corresponding terms of the geometric progression 2, 4, 8, 16, 32 and 128, 32, 8, 2, $\frac{1}{2}$

Sol :

Sum of the product of corresponding terms of the G.M.s

2,4,8,16,32 and $128,32,8,2, \frac{1}{2}$

$=2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \frac{1}{2}$

=256+128+64+32+16

Here, $a=256, r=\frac{128}{256}=\frac{1}{2}, n=5$

$S_{5}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{256\left(1-\frac{1}{2}\right)^{5}}{1-\frac{1}{2}}$

$\mathrm{S}_{5}=\frac{256\left(1-\frac{1}{32}\right)}{\frac{1}{2}}=\frac{256 \times 2}{1}\left(\frac{31}{32}\right)$

$S_{5}=8 \times 2 \times 31=496$

Question 18

Evaluate $\sum_{n=1}^{50}\left(2^{n}-1\right)$

Sol :

$\sum_{n=1}^{50}\left(2^{n}-1\right)$

Here n = 1, 2, 3,….,50

$\mathrm{S}_{n}=2^{1}-1+2^{2}-1+2^{3}-1+2^{4}-1 \ldots 2^{50}-1$

$\mathrm{S}_{n}=\left(2^{1}+2^{2}+2^{3}+2^{4} \ldots 2^{50}\right)-1 \times 50$

$\mathrm{~S}_{n}=2+4+8+16 \ldots 2^{50}-50$

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}-50=\frac{2\left(2^{50}-1\right)}{2-1}-50$

$\mathrm{~S}_{n}=2 \times 2^{50}-2-50$

$\mathrm{~S}_{n}=2^{51}-52$

Question 19

Find the sum of n terms of a series whose mth term is $2^{m}+2 m$.

Sol :

$a_{m}=2^{m}+2 m$

$a_{1}=2^{1}+2 \times 1=2+2$

$a_{2}=2^{2}+2 \times 2=4+4$

$a_{3}=2^{3}+2 \times 3=8+6$

$a_{4}=2^{4}+2 \times 4=16+8$

$a_{n}=2^{n}+2 \times n$

Sum $=(2+4+8+16 \ldots \text{ n term })+(2+4+6+8 \ldots \text{ n term })$

$\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}+\frac{n}{2}[2 a+(n-1) d]$

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

$\{\because a=2, r=2, d=2\}$

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

$=2\left(2^{n}-1\right)+n(2+n-1)$

$=2\left(2^{n}-1\right)+n(n+1)$

Question 20

Sum the series

$x(x+y)+x^{2}\left(x^{2}+y^{2}\right)+x^{3}\left(x^{3}+y^{3}\right)$… to n terms.

Sol :

Given

$S_{n}=x(x+y)+x^{2}\left(x^{2}+y^{2}\right)+x^{3}\left(x^{3}+y^{3}\right) \ldots$n terms

$\mathrm{S}_{n}=x^{2}+x y+x^{4}+x^{2} y^{2}+x^{6}+x^{3} y^{3}+\ldots$n terms

$\mathrm{S}_{n}=x^{2}+x^{4}+x^{6}+\ldots n$ terms $+x y+x^{2} y^{2}+x^{3} y^{3}+\ldots$

$\mathrm{S}_{n}=\frac{x^{2}\left[\left(x^{2}\right)^{n}-1\right]}{x^{2}-1}+\frac{x y\left[(x y)^{n}-1\right]}{x y-1}$

$\left\{\begin{array}{l}\text { In the first G.P. } a=x^{2}, r=x^{2} \\ \text { In second G.P. } a=x y, r=x y\end{array}\right\}$

$S_{n}=\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1}+\frac{x y\left[(x y)^{n}-1\right]}{x y-1}$

Question 21

Find the sum of the series

$1+(1+x)+\left(1+x+x^{2}\right)+\ldots$ to n terms, $x \neq 1$

Sol :

$1+(1+x)+\left(1+x+x^{2}\right)+\ldots$ n terms, $x \neq 1$

Multiply and divide by (1 – x)

$\Rightarrow \frac{(1-x)}{(1-x)}+\frac{(1-x)(1+x)}{(1-x)}+\frac{(1-x)\left(1+x+x^{2}\right)}{(1-x)}+\ldots$ upto $n$ terms

$\Rightarrow \frac{1}{(1-x)}\left[(1-x)+(1+x)^{2}+\left(1+x^{3}\right)+\ldots\right.$ upto n terms

$\Rightarrow \frac{1}{(1-x)}[(1+1+1+\ldots$ upto $n$ terms $)-(x+x^{2}+x^{3}+\ldots$ upto n terms )

$\Rightarrow \frac{1}{(1-x)}\left[n-\frac{x\left(1-x^{n}\right)}{(1-x)}\right]$

$\left[\because\right.$ Using $\mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$ as $\left.r<1\right]$

$\Rightarrow \frac{1}{(1-x)}\left[\frac{n(1-x)-x\left(1-x^{n}\right)}{(1-x)}\right]$

$\Rightarrow \frac{1}{(1-x)^{2}}\left[n(1-x)-x\left(1-x^{n}\right)\right]$

Question 22

Find the sum of the following series to n terms:

(i) 7 + 77 + 777 + …

(ii) 8 + 88 + 888 + …

(iii) 0.5 + 0.55 + 0.555 + …

Sol :

(i) 7 + 77 + 777 + … n terms

= 7[1 + 11 + 111 + … n terms]

$=\frac{7}{9}[9+99+999+\ldots n$ terms $]$

$=\frac{7}{9}[(10-1)+(100-1)+(1000-1)+\ldots$ n terms]

$=\frac{7}{9}[10+100+1000+\ldots n$ terms $-(1+1+1...$ n terms ]

$=\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]$

$=\frac{7}{9}\left[\frac{10 \times 10^{n}-10}{9}-n\right]$

$=\frac{7}{81}\left[10^{n+1}-10-9 n\right]$

$=\frac{7}{81}\left[10^{n+1}-9 n-10\right]$

(ii) $8+88+888+\ldots n$ terms

$=8[1+11+111+\ldots n$ terms $]$

$=\frac{8}{9}[9+99+999+\ldots n$ terms $]$

$=\frac{8}{9}[(10-1)+(100-1)+(1000-1)+...$n terms]

$=\frac{8}{9}[10+100+1000+\ldots n$ terms $-1 \times n]$

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

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

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

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

(iii) 0.5+0.55+0.555+... n terms

=5[0.1+0.11+0.111+... n terms ]

$=\frac{5}{9}[0.9+0.99+0.999+\ldots$ n terms ]

$=\frac{5}{9}[(1-0.1)+(1-0.01)+(1-0.001)+...$ n terms)

$=\frac{5}{9}[1+1+1+\ldots n$ terms $-(0.1+0.01+0.001)+...$n terms]

$=\frac{5}{9}\left[n-\frac{0.1\left[1-(-0.1)^{n}\right.}{1-0.1}\right]$

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

$=\frac{5}{81}\left[9 n-1+\frac{1}{10^{n}}\right]$