Question

if $S=\frac{1}{3}-\frac{13}{3}+\frac{169}{3}-.....\infty$, then find the value of $14S$

Answer 1

The given sequence is on infinite GP whose first term is $a=\frac{1}{3}$

Common Ratio $r=\frac{-13}{3}\times 3=-13$

$⟹S=\frac{a}{1-r}=\frac{\frac{1}{3}}{1+13}$

$⟹14S=\frac{1}{3}$