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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum of $n$ terms of the series $0.6+0.66+0.666+......$

$\begin{array}{1 1}\large\frac{2n}{3}-\frac{(10^n-1)}{27.10^n} \\\large\frac{2n}{3}-\frac{2(10^n-1)}{27.10^n} \\ \large\frac{2n}{3}-\frac{2(10^n-1)}{9.10^n} \\\large\frac{2n}{3}-\frac{(10^n-1)}{9.10^n} \end{array} $

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  • Sum of $n$ terms of a G.P is $a.\large\frac{1-r^n}{1-r}$
Given series is $S_n=0.6+0.66+0.666+......$
Taking $6$ common we get
$S_n=6(0.1+0.11+0.111+..........)$
Multiplying and dividing by $9$ we get
$S_n=\large\frac{6}{9}$$(0.9+0.99+0.999+..........)$
$\qquad =\large\frac{2}{3}$$\big[(1-0.1)+(1-0.01)+(1-0.001)+..........\big]$
$\qquad =\large\frac{2}{3}$$\big[(1+1+1+............)-(0.1+0.01+0.001+........)\big]$
But $ 1+1+1+........=n$ and
$0.1+0.01+0.001+.........$ is a G.P. with $1^{st}$ term $=a=0.1$ and common ratio $=r=0.1$
We know that sum of $n$ terms of a G.P is $a.\large\frac{1-r^n}{1-r}$
$\therefore \:\:0.1+0.01+0.001+.......=0.1\times \large\frac{1-(0.1)^n}{1-0.1}$
$1-(0.1)^n=1-\large\frac{1}{10^n}=\frac{10^n-1}{10^n}$
$1-0.1=1-\large\frac{1}{10}=\frac{9}{10}$
$\therefore\:0.1+0.01+0.001+.......=\large\frac{1}{10}\times \large\frac{10^n-1}{10^n}\times \frac{10}{9}$
$\qquad\qquad\:\:=\large\frac{10^n-1}{9.10^n}$
Substituting the values in $S_n$ we get
$S_n=\large\frac{2}{3}$$\big[n-\large\frac{10^n-1}{9.10^n}\big]$
$\qquad\:=\large\frac{2n}{3}-\frac{2(10^n-1)}{27.10^n}$
answered Apr 3, 2014 by rvidyagovindarajan_1
 

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