Question

In the geometric series $2+4+8+....$, starting from the first term how many consecutive terms are needed to yield the sum $1022$?

Answer 1

Given the geometric series is $2+4+8+...$
Let $n$ be the number of terms required to get the sum.
Here $a=2,r=2,S_n=1022$.
To find $n$, let us consider
$S_n=\frac{a[r^n-1]}{r-1}$ if $r\ne 1$
$=\left(2\right)\left[\frac{2^n-1}{1}\right]=2(2^n-1)$.
But $S_n=1022$ and hence $2(2^n-1)=1022$
$\Rightarrow 2^n-1=511$
$\Rightarrow 2^n=512=2^9$

Thus, $n=9$