Question

i The sum to infinity of the GP -3,1,-1/3,1/9... is​

Answer 1

Given

$\sf \to\: - 3,1, \dfrac{ - 1}{3} , \dfrac{1}{9} ,$

$\sf \to \: first \: term \: (a) = - 3$

$\sf \to \: commom \: ratio(r) = \dfrac{a_{2}}{a_1}$

$\sf \to \: commom \: ratio(r) = \dfrac{ - 1}{3}$

Formula

$\sf \to \: S_ \infty = \dfrac{a}{1 - r}$

Now Put the value on formula

$\sf \to \: S_ \infty = \dfrac{ - 3}{1 - \bigg(\dfrac{ - 1}{3} \bigg) }$

$\sf \to \: S_ \infty = \dfrac{ - 3}{1 + \dfrac{ 1}{3}}$

$\sf \to \: S_ \infty = \dfrac{ - 3}{ \dfrac{ 3 + 1}{3}}$

$\sf \to \: S_ \infty = \dfrac{ - 3}{ \dfrac{ 4}{3}}$

$\sf \to \: S_ \infty = \dfrac{ - 3 \times 3}{ 4}$

$\sf \to \: S_ \infty = \dfrac{ - 9}{ 4}$

Answer

$\sf \to \: S_ \infty = \dfrac{ - 9}{ 4}$