Question

Sum of $n$ terms of the series $8+88+888+.$ equals

• A. $\frac{8}{18}[{10}^{n+1}-9n-10]$
• B. $\frac{8}{18}[{10}^n-9n-10]$
• C. $\frac{8}{18}[{10}^{n+1}+9n-10]$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{8}{18}[{10}^{n+1}-9n-10]$

series $=8+88+888+...$

as$\frac{a_2}{a_1}=\frac{88}{8}=11$

and $\frac{a_3}{a_2}=\frac{888}{88}=51$

$\therefore \frac{a_2}{a_1}\ne \frac{a_3}{a_2}.$ This is not a G.P.

But, $S=8+88+888+..$

$=8(1+11+111+...)$

$\frac{8}{9}\left(9\right(1+11+111+..)$

$\frac{8}{9}(9+99+999+...)$

$\frac{8}{9}\left(\right(10-1)+({10}^2-1)+({10}^3-1)+...)$

$\Rightarrow \frac{8}{9}(10+{10}^2+{10}^3...)-(1+1+1+..)$

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

$S\Rightarrow \frac{8}{9}\left(\frac{10({10}^n-1)-9n}{9}\right)$

$\frac{8}{9}\left(\frac{{10}^{n+1}-10-9n}{9}\right)$ [G.P with A = 10, B = 10 $S=\frac{A(R^n-1)}{R-1}]$

$\Rightarrow \frac{8}{81}({10}^{n+1}-9n-10)$