Discussion Forum

 The sum of first 20 terms of the sequence 0.7 ,0.77,0.777, .... is 

Answer:

Explanation:

 Let S=0.7+0.77+0.777+.....=  $\frac{7}{10}+\frac{77}{10^{2}}+\frac{777}{10^{3}}+....$  + up to 20 terms

$=7\left[\frac{1}{10}+\frac{11}{10^{2}}+\frac{111}{10^{3}}+....upto 20 terms \right]$

   $=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{10^{2}}+\frac{999}{10^{3}}+....upto 20 terms \right]$

  $=\frac{7}{9}\left[(1-\frac{1}{10})+(1-\frac{1}{10^{2}})+(1-\frac{1}{10^{3}})+....upto 20 terms \right]$

$=\frac{7}{9}\left[(1+1+......+upto 20 terms)-(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+....upto 20 terms) \right]$

$=\frac{7}{9}\left[20-\frac{\frac{1}{10}\left\{{1-(\frac{1}{10})^{20}}\right\}}{1-\frac{1}{10}}\right]$

[   $\therefore$ $\sum_{i=1}^{20}=20$  and sum of n terms of GPS   $s_{n}=\frac{a(1-r^{n})}{1-r}$ when (r<1)]

   $=\frac{7}{9}\left[20-\frac{1}{9}\left\{1-(\frac{1}{10})^{20}\right\}\right]$

 $=\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}(\frac{1}{10})^{20}\right]$

 $=\frac{7}{81}[179+(10)^{-20}]$