Find the sum of the sequence $7,77,777,7777, \ldots$ to $n$ terms.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

This is not a $G.P.,$ however, we can relate it to a $G.P.$ by writing the terms as

${S_n} = 7 + 77 + 777 + 7777 +  \ldots {\rm{ }}$ to $ n $ terms 

$ = \frac{7}{9}[9 + 99 + 999 + 9999 +  \ldots $ to $ n $ term $]$

$ = \frac{7}{9}[(10 - 1) + \left( {{{10}^2} - 1} \right) + \left( {{{10}^3} - 1} \right) + \left( {{{10}^4} - 1} \right) +  \ldots n{\rm{ }}$ term $]$

$=\frac{7}{9}[(10+10^{2}+10^{3}+\ldots n \text { terms })$

$-(1+1+1+\ldots n \text { terms })]$

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

Similar Questions

If $a,\;b,\;c,\;d$ and $p$ are different real numbers such that $({a^2} + {b^2} + {c^2}){p^2} - 2(ab + bc + cd)p + ({b^2} + {c^2} + {d^2}) \le 0$, then $a,\;b,\;c,\;d$ are in

  • [IIT 1987]

The greatest integer less than or equal to the sum of first $100$ terms of the sequence $\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots \ldots$ is equal to

  • [JEE MAIN 2022]

Find the $20^{\text {th }}$ and $n^{\text {th }}$ terms of the $G.P.$ $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$

Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$

The first and last terms of a $G.P.$ are $a$ and $l$ respectively; $r$ being its common ratio; then the number of terms in this $G.P.$ is