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Question Number 126253 by mohammad17 last updated on 18/Dec/20

Answered by physicstutes last updated on 18/Dec/20

$$\mathrm{Exactly}$$$$a+ar+{ar}^2+...+{ar}^{n-1}=\underset{r=1}{\sum ^n}{ar}^{n-1}=\frac{a(1-r^n)}{1-r}$$$$\mathrm{for}\mathrm{convergence}\underset{n\to \mathrm{\infty }}{\lim }{S}_n=\underset{n\to \mathrm{\infty }}{\lim }\underset{r=1}{\sum ^n}{ar}^{n-1}$$$$\Rightarrow S_{\mathrm{\infty }}=\underset{n\to \mathrm{\infty }}{\lim }\left[\frac{a(1-r^n)}{1-r}\right]$$$$\mathrm{if}\mid r\mid <1\mathrm{then}\mathrm{as}n\to \mathrm{\infty },r^n\to 0$$$$\Rightarrow S_{\mathrm{\infty }}=\frac{a}{1-r}.$$$$\mathrm{notice}\mathrm{if}\mid r\mid >1,r^n\to \mathrm{\infty }\mathrm{as}r\to \mathrm{\infty }$$$$\Rightarrow S_{\mathrm{\infty }}=\mathrm{\infty }\mathrm{thus}\mathrm{for}\mathrm{convergence},\mathrm{of}{S}_n,\mid r\mid <1$$

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