The sum of an infinite geometric series is 3. | vedclass

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Question

(a) ${({S_1})_\infty } = \frac{a}{{1 - r}} = 3$ or $a = 3\,(\,1 - r)$ &hellip;..$(i)$

${({S_2})_\infty } = \frac{{{a^2}}}{{1 - {r^2}}} = 3$

or $

Vedclass · Accepted Answer

$\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{{16}},.....$

Vedclass · Answer

(a) ${({S_1})_\infty } = \frac{a}{{1 - r}} = 3$ or $a = 3\,(\,1 - r)$ &hellip;..$(i)$

${({S_2})_\infty } = \frac{{{a^2}}}{{1 - {r^2}}} = 3$

or ${a^2} = 3\,(1 - {r^2})$ or $9\,{(1 - r)^2} = 3\,(1 - {r^2})$ [by $(i)$]

or $3\,(1 - 2r + {r^2}) = 1 - {r^2}$ or $2{r^2} - 3r + 1 = 0$

or $(r - 1)\,(2r - 1) = 0$,

$	herefore $$r = 1,\frac{1}{2}$

If $r = 1,$ then $a = 3(1 - 1) = 0$ which is impossible.

If $r = \frac{1}{2},$then $a = 3\,\left( {1 - \frac{1}{2}} ight) = 3/2$

So first series is $3/2, 3/4, 3/8, 3/16,.....$

The sum of an infinite geometric series is $3$. A series, which is formed by squares of its terms, have the sum also $3$. First series will be

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