Question

How many terms of the G.P. 3, $\frac{3}{2},\frac{3}{4},....$ be taken together to make $\frac{3069}{512}$ ?

Answer 1

$3,\frac{3}{2},\frac{3}{4},....isaG.P.$

and $S_n=\frac{3069}{512},a=3,r=\frac{1}{2}$

$S_n=\frac{a(1-r^n)}{1-r}$

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

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

${1023.2}^n={1024.2}^n-1024$

$1024=2^n$

$\Rightarrow 2^{10}=2^n$

$\Rightarrow n=10$