Show that in an infinite G.P. with common ratio r (|r|<1 ), each terms bears a constant ratio to the sum of all terms that follow it.

Let a be first term and r be common ratio of G.P. Here, a_n (a_ n+1 +a_ n+2 + ) = ar^ n-1 ar^ n +ar^ n+1 + = ar^ n-1 ar^ n (1+r+r^2+ ) = ar^ n-1 ar^ n (1 1-r ) = ( 1-r r ) Since r is a constant, so ( a_n a_ n+1 +a_ n+2 + )=k (constant) Such that k= ( 1-r r )

Show that in an infinite G.P. with common ratio r (|r|<1 ), each …

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Question

Show that in an infinite G.P. with common ratio $\text{r}\big(|\text{r}|<1\big),$ each terms bears a constant ratio to the sum of all terms that follow it.

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Answer

Let a be first term and r be common ratio of G.P.
Here, $\frac{\text{a}_\text{n}}{\big(\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty\big)}=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}+\text{ar}^{\text{n}+1}+\ \dots}$
$=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\big(1+\text{r}+\text{r}^2+\ \dots\infty\big)}$
$=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\Big(\frac{1}{1-\text{r}}\Big)}$
$=\Big(\frac{1-\text{r}}{\text{r}}\Big)$
Since r is a constant, so
$\Big(\frac{\text{a}_\text{n}}{\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty}\Big)=\text{k}\ (\text{constant})$
Such that $\text{k}=\Big(\frac{1-\text{r}}{\text{r}}\Big)$