The sum of few terms of any ratio series is 7 | vedclass

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Question

(a) $	herefore {n^{th}}$ term of series $ = a{r^{n - 1}}$$ = a\,{(3)^{n - 1}} = 486$ &hellip;..$(i)$

and sum of $n$ terms of series.

${S_n} = \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a) $	herefore {n^{th}}$ term of series $ = a{r^{n - 1}}$$ = a\,{(3)^{n - 1}} = 486$ &hellip;..$(i)$

and sum of $n$ terms of series.

${S_n} = \frac{{a({3^n} - 1)}}{{3 - 1}}$ &hellip;..$(ii)$

From $(i),$ $a\left( {\frac{{{3^n}}}{3}} ight) = 486$ or $a{.3^n} = 3 	imes 486 = 1458$

From $(ii),$ $a{.3^n} - a = 728 	imes 2$ or $a{.3^n} - a = 1456$

$1458 - a = 1456$

$&rArr;$  $a = 2$.

The sum of few terms of any ratio series is $728$, if common ratio is $3$ and last term is $486$, then first term of series will be

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