The sum to infinity of the progression 9 - 3 | vedclass

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Question

(C) The given series $9 - 3 + 1 - \frac{1}{3} + .....$ is a Geometric Progression $(G.P.)$.Here,the first term $a = 9$.The common ratio $r = \frac{-3}

Vedclass · Accepted Answer

$27/4$

Vedclass · Answer

(C) The given series $9 - 3 + 1 - \frac{1}{3} + .....$ is a Geometric Progression $(G.P.)$.Here,the first term $a = 9$.The common ratio $r = \frac{-3}{9} = -\frac{1}{3}$.The sum to infinity of a $G.P.$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$,provided $|r| Substituting the values,we get $S_{\infty} = \frac{9}{1 - (-1/3)} = \frac{9}{1 + 1/3} = \frac{9}{4/3} = \frac{9 	imes 3}{4} = \frac{27}{4}$.

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + .....$ is

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