Exercise 2.8 - Chapter 2 - Numbers and Sequences - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Ex $2.8$
Question 1.
Find the sum of first $n$ terms of the G.P.
(i) $5,-3, \frac{9}{5},-\frac{27}{25}, \ldots \ldots \ldots$.
(ii) $256,64,16, \ldots \ldots .$
Solution:
(i) $5,-3, \frac{9}{5}, \frac{-27}{25}$,
Here $a=5, r=\frac{t_{2}}{t_{1}}=\frac{-3}{5}<1$
$\begin{aligned}
\mathrm{S}_{n} &=a\left(\frac{1-r^{n}}{1-r}\right) \\
=& 5\left[\frac{1-\left(\frac{-3}{5}\right)^{n}}{1-\left(\frac{-3}{5}\right)}\right]=5\left[\frac{1-\left(\frac{-3}{5}\right)^{n}}{1-\frac{-3}{5}}\right]
\end{aligned}$
$=5 \frac{\left[1-\left(\frac{-3}{5}\right)^{n}\right]}{\frac{8}{5}}=5 \times \frac{5}{8}\left(1-\left(\frac{-3}{5}\right)^{n}\right)$
$\therefore S_{n}=\frac{25}{8}\left(1-\left(\frac{-3}{5}\right)^{n}\right)$
(ii) $256,64,16, \ldots \ldots$

$\begin{aligned}
\mathrm{a}=256 \\
r=\frac{64}{256}=\frac{4}{16} &=\frac{1}{4}<1 \\
\mathrm{~S}_{n} &=\frac{a\left(1-r^{n}\right)}{1-r} \\
&=256\left(\frac{1-\left(\frac{1}{4}\right)^{n}}{\left.1-\frac{1}{4}\right)}\right.\\
&=\frac{1024}{3}\left(1-\left(\frac{1}{4}\right)^{n}\right) \\
&\left.=\frac{256}{3}\right) \\
&\left.=\frac{3}{4}\right)
\end{aligned}$

Question $2 .$
Find the sum of first six terms of the G.P. $5,15,45, \ldots$
Solution:

G.P. $=5,15,45,15$
$n=6, a=5, r=\frac{15}{5}=3>1$
$\therefore \quad \mathrm{S}_{n}=a \frac{\left(r^{n}-1\right)}{r-1}$
$\mathrm{S}_{6}=5\left(\frac{3^{6}-1}{3-1}\right)$
$=5 \frac{\left(3^{6}-1\right)}{2}=\frac{5}{2}(729-1)$
$=\frac{5}{2} \times 728=5 \times 364$
$=1820$

Question $3 .$
Find the first term of the GP. whose common ratio 5 and whose sum to first 6 terms is 46872 .
Solution:
Common ratio, $\mathrm{r}=5$
$\mathrm{S}_{6}=46872$
$\therefore \quad \frac{a\left(r^{6}-1\right)}{r-1}=46872$
$\Rightarrow \quad \mathrm{S}_{6}=\frac{a\left(5^{6}-1\right)}{5-1}=46872$
$\Rightarrow \quad a=\frac{46872 \times 4}{[25 \times 25 \times 25-1]}$
$\Rightarrow \quad a=\frac{187488}{15624}=12$
$\Rightarrow \quad$ first term $=12$

Question $4 .$
Find the sum to infinity of (i) $9+3+1+\ldots$.
(ii) $21+14+\frac{28}{3}+\ldots .$
Solution:
(i) $9+3+1+\ldots$
$\begin{aligned}
&a=9, r=\frac{3}{9}=\frac{1}{3}<1 \\
&\mathrm{~S}_{\infty}=\frac{a}{1-r}=\frac{9}{1-\frac{1}{3}} \\
&=\frac{\frac{9}{3-1}}{3}=\frac{9}{3}=9 \times \frac{3}{2}=\frac{27}{2}
\end{aligned}$
(ii) $21+14+\frac{28}{3}+\ldots$
Here $a=21, r=\frac{14}{21}=\frac{2}{3}$
$\begin{aligned}
S_{\infty} &=\frac{a}{1-r}=\frac{21}{1-\frac{2}{3}}=\frac{21}{\frac{3-2}{3}} \\
&=\frac{21}{\frac{1}{3}}=21 \times 3=63
\end{aligned}$

$\mathrm{S}_{\infty}=63$

Question $5 .$
If the first term of an infinite G.P. is 8 and its sum to infinity is $\frac{32}{3}$ then find the common ratio.

Solution:

$\begin{aligned}
\mathrm{a}=8 & \\
\mathrm{~S}_{\infty}=\frac{32}{3} \Rightarrow \frac{a}{1-r} &=\frac{32}{3} \\
\frac{8}{1-r} &=\frac{32}{3} \\
32(1-r) &=\frac{24}{24}=\frac{6}{8}=\frac{3}{4} \\
1-r=\frac{2}{32}=\frac{3}{4}-1 &=\frac{3-4}{4} \\
-r=r &=\frac{-1}{4} \Rightarrow r=\frac{1}{4}
\end{aligned}$

Question 6.
Find the sum to $n$ terms of the series
(i) $0.4+0.44+0.444+\ldots .$ to $n$ terms
(ii) $3+33+333+\ldots$ to $n$ terms
Solution:
(i) $0.4+0.44+0.444+\ldots$ to $\mathrm{n}$ terms $=4(0.1+0.11+0.111+\ldots$ to $\mathrm{n}$ terms $)$ $=\frac{4}{9}(0.9+0.99+0.999+\ldots$ to $\mathrm{n}$ terms $)$

$\begin{aligned}
&\left.=\frac{4}{9}(1-0.1)+(1-0.01)+(1-0.001)+\ldots n \text { terms }\right)\\
&=\frac{4}{9}(1+1+1+\ldots n \text { terms })-\left(0.1+0.1^{2}+0.1^{3}+\right.\\
&=\frac{4}{9}\left[n-\frac{1}{9}\left[1-\left(\frac{1}{10}\right)^{n}\right]\right.\\
&=\frac{4}{9} n-\frac{4}{81}\left(1-\left(\frac{1}{10}\right)^{n}\right]\\
&\text { (ii) } 3+33+333+\ldots \text { to } n \text { terms. }\\
&=3(1+11+111+\text { to } n \text { terms })\\
&=\frac{3}{9}(9+99+999+\ldots \text { to } n \text { terms })\\
&\frac{1}{3}=[(10-1)+(100-1)+(1000-1)+\ldots \text { to } n \text { terms }]\\
&=\frac{1}{3}[(10+100+1000+. .)+(-1) n]
\end{aligned}$

$=\frac{1}{3}\left(\frac{10\left(10^{n}-1\right)}{9}-n\right)$
$=\frac{10}{27}\left(10^{n}-1\right)-\frac{n}{3}$

Question 7.
Find the sum of the Geometric series $3+6+12+\ldots \ldots+1536 .$
Solution:
$3+6+12+\ldots \ldots+1536$ Here $a=3$

$\begin{aligned} r &=\frac{6}{3} \quad=2>1 \\ \mathrm{~S}_{n} &=\frac{a\left(r^{n}-1\right)}{(r-1)} \\ \mathrm{S}_{10} &=\frac{3\left(2^{10}-1\right)}{2-1} \\ &=3(1024-1) \\ &=3 \times 1023=3069 \end{aligned}$

Question $8 .$
Kumar 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 the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs $\sqcup 2$ to mail one letter, find the amount spent on postage when 8th set of letters is mailed.

Solution:
Kumar (1) $2 \times 4$
Cost of posting $1^{\text {st }}$ set of Letters.
Cost of posting $2^{\text {nd }}$ set of letters

Question $9 .$
Find the rational form of the number $0 . \overline{123}$
Solution:
Let $x=0.123123123 \ldots \ldots . . \Rightarrow x=0 . \overline{123} \ldots . .(1)$
Multiplying 1000 on both rides
$1000 \mathrm{x}=123.123123 \ldots \Rightarrow 1000 \mathrm{x}=123 . \overline{123}$
(2) $-(1)=1000 x-x 123 . \overline{123}-0 . \overline{123}$.
$\Rightarrow 999 x=123$
$\begin{aligned}
&\Rightarrow x=\frac{123}{999} \\
&\Rightarrow x=\frac{41}{333} \text { Rational number }
\end{aligned}$

Question $10 .$
If $S n=(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots n$ terms then prove that $(x-y)$ $S_{n}=\left[\frac{x^{2}\left(x^{n}-1\right)}{x-1}-\frac{y^{2}\left(y^{n}-1\right)}{y-1}\right]$
Solution:
$\begin{aligned}
&\text { Sn }=(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots n \text { terms } \\
&\Rightarrow x . S_{n}=(x+y) x+\left(x^{2}+x y+y^{2}\right) x+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right) x+\ldots \\
&\Rightarrow x . S_{n}=x^{2}+x y+x^{3}+x^{2} y+y^{2} x+x^{4}+x^{3} y \\
&+x^{2} y^{2}+y^{3} x+\ldots \ldots . .(1)
\end{aligned}$

$\begin{aligned}
&=y . \mathrm{S}_{\mathrm{n}}=(\mathrm{x}+\mathrm{y}) \mathrm{y}+\left(\mathrm{x}^{2}+\mathrm{xy}+\mathrm{y}^{2}\right) \mathrm{y}+ \\
&\left(\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{y}+\mathrm{xy} \mathrm{y}^{2}+\mathrm{y}^{3}\right) \mathrm{y}+\ldots . \\
&=\mathrm{y} . \mathrm{S}_{\mathrm{n}}=\mathrm{xy}+\mathrm{y}^{2}+\mathrm{x}^{2} \mathrm{y}+\mathrm{xy} \mathrm{y}^{2}+\mathrm{y}^{3}+\mathrm{x}^{3} \mathrm{y} \\
&+\mathrm{x}^{3} \mathrm{y}+\mathrm{x}^{2} \mathrm{y}^{2}+\mathrm{xy}^{3}+\mathrm{y}^{4}+ \\
&(1)-(2) \Rightarrow \\
&x \mathrm{~S}_{n}-y \mathrm{~S}_{n}=\left(x^{2}+\not x+x^{3}+y^{2} y+y^{2} y+x^{4}+x^{3} y\right. \\
&\left.+x^{2} y^{2}+y^{2} x+\ldots . .\right)-\left(x y+y^{2}+x^{2} y+x^{2}+y^{3}+\right. \\
&\Rightarrow(x-y) \mathrm{S}_{n}=\left(x^{2}+x^{3}+x^{4}+\ldots y^{4}+\ldots .\right. \\
&=\frac{\left.x^{2}+y^{3}+y^{4}+\ldots .\right)}{x-1}-\frac{y^{2}\left(y^{n}-1\right)}{y-1} \\
&\Rightarrow(x-y) \mathrm{S}_{n}=\left[\frac{x^{2}\left(x^{n}-1\right)}{x-1}-\frac{y^{2}\left(y^{n}-1\right)}{y-1}\right]
\end{aligned}$
Hence proved.