Question

The sum of $n$ terms of a geometric series is $255$. If the last term is $128$, the common ratio is $2$, find $n$.

Answer 1

$\frac{A(R^n-1)}{R-1}=255$

$R-1$

$A(2^n-1)=255$

$A{R}^{n-1}=128$
$A=\frac{128}{2^{n-1}}$

$\frac{128}{2^{n-1}}(2^n-1)=255$

$\frac{128\times 2}{2^n}(2^n-1)=255$

$(1-2^{-n})=\frac{255}{256}$

$1-\frac{255}{256}=2^{-n}$

$\frac{1}{256}=\frac{1}{2^n}$

$=2^n=256=2^8$

$n=8$