Question

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

Answer 1

Given : sequence $7,77,777,7777,...$ upto $n$ terms
Here, $\frac{77}{7}=11$
and $\frac{777}{77}=10.09$
$\because$ Common ratio is not same.
$\therefore$ The given sequence is not G.P

We need to find sum $=7+77+777+7777+\cdots$ upto $n$ terms
$=7(1+11+111+\cdots$ upto $n$ terms)
Multiplying & dividing by $9$
$=\frac{7}{9}\left[9\right(1+11+111+...$ upto $n$ terms)]
$=\frac{7}{9}[9+99+999+9999+...$ upto $n$ terms]
$=\frac{7}{9}\left[\right(10-1)+(100-1)+(1000-1)+...$ upto $n$ terms]
$=\frac{7}{9}\left[\right(10+100+1000+...n\text{terms})-(1+1+1+...\text{upto}n\text{terms}]$
$\text{Sum}=\frac{7}{9}\left[\right(10+100+1000+...\text{terms})-n\times 1]\cdots \left(i\right)$

Now, $10+100+1000+...n\text{terms is a G.P}$
Here, $a=10$ and $r=10>1$
We know,
sum of $n$ terms of G.P., $=S_n=\frac{a(r^n-1)}{r-1};r>1$
$S_n=\frac{10({10}^n-1)}{10-1}$
$\Rightarrow S_n=\frac{10({10}^n-1)}{9}$

Now substituting this value in $\left(i\right),$ we get
Sum $=\frac{7}{9}[\frac{10({10}^n-1)}{9}-n]$