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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum to $n$ terms of the sequence $8,88,888,.............$

$\begin{array}{1 1}\large\frac{8}{9}[\large\frac{10}{9}(10^n-1)] \\\large\frac{8}{9}[\large\frac{10}{9}(10^n-1)-1] \\ \large\frac{8}{9}[\large\frac{10}{9}(10^n-1)-n] \\ \large\frac{8}{9}[\large\frac{10}{9}10^n-1-n] \end{array} $

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  • Sum of $n$ terms of a G.P=$a.\large\frac{r^n-1}{r-1}$
Given series $S=8+88+888+........$
Taking $8$ common
$\Rightarrow \:S=8(1+11+111+.........)$
Multiplying and dividing by 9
$S=\large\frac{8}{9}$$(9+99+999+..............)$
Splitting each term
$\Rightarrow\:S=\large\frac{8}{9}$$[(10-1)+(100-1)+(1000-1)+.........]$
$S=\large\frac{8}{9}$$[(10+100+1000+......)-(1+1+1+..........)]$
Step 2
$10+100+1000+.........$ is a G.P. with $a=10\:\:and\:\:r=10$
We know that sum of $n$ terms of a G.P=$a.\large\frac{r^n-1}{r-1}$
$\therefore\:10+10^2+10^3+.....=10.\large\frac{10^n-1}{10-1}=\frac{10}{9}$$(10^n-1)$
and
$1+1+1+.......=n$
$\Rightarrow\:S=\large\frac{8}{9}$$[\large\frac{10}{9}$$(10^n-1)-n]$
answered Feb 28, 2014 by rvidyagovindarajan_1
 

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