Question

How many terms of the G.P. $3,\frac{3}{2},\frac{3}{4}.....$ are needed to give the sum $\frac{3069}{512}$ ?

Answer 1

Here, a = 3 and common ratio $\frac{1}{2}$

$S_n=\frac{3069}{512}=\frac{3(1-\frac{1}{2^n})}{\frac{1}{2}}$

$1-\frac{1}{2^n}=\frac{3069}{512\times 6}=\frac{1023}{512\times 2}$

$1-\frac{1023}{1024}=\frac{1}{2^n}$

$\frac{1}{2^n}=\frac{1}{1024}$

$n=10$