The sum to infinity of a geometric progressio | vedclass

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Question

(A) The sum to infinity $(S_{\infty})$ of a geometric progression is given by the formula: $S_{\infty} = \frac{a}{1 - r}$,where $a$ is the first term

Vedclass · Accepted Answer

$7/16$

Vedclass · Answer

(A) The sum to infinity $(S_{\infty})$ of a geometric progression is given by the formula: $S_{\infty} = \frac{a}{1 - r}$,where $a$ is the first term and $r$ is the common ratio.Given: $S_{\infty} = 4/3$ and $a = 3/4$.Substituting the values into the formula: $\frac{4}{3} = \frac{3/4}{1 - r}$.Multiplying both sides by $(1 - r)$ and rearranging: $1 - r = \frac{3/4}{4/3} = \frac{3}{4} 	imes \frac{3}{4} = \frac{9}{16}$.Therefore,$r = 1 - \frac{9}{16} = \frac{16 - 9}{16} = 7/16$.

The sum to infinity of a geometric progression is $4/3$ and the first term is $3/4$. The common ratio is

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