In a geometric progression,if the ratio of th | vedclass

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(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.The sum of the first $5$ terms is $S_5 = a + ar + ar^2 + ar^3 + ar^4 = \frac{a(r^5 - 1)

Vedclass · Accepted Answer

$28$

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(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.The sum of the first $5$ terms is $S_5 = a + ar + ar^2 + ar^3 + ar^4 = \frac{a(r^5 - 1)}{r - 1}$.The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.The sum of these reciprocals is $S'_5 = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4} = \frac{1}{a} \left( \frac{1 - (1/r)^5}{1 - 1/r} \right) = \frac{1}{a} \left( \frac{(r^5 - 1)/r^5}{(r - 1)/r} \right) = \frac{1}{a} \cdot \frac{r^5 - 1}{r^5} \cdot \frac{r}{r - 1} = \frac{r^5 - 1}{a r^4 (r - 1)}$.Given the ratio $\frac{S_5}{S'_5} = 49$:$\frac{\frac{a(r^5 - 1)}{r - 1}}{\frac{r^5 - 1}{a r^4 (r - 1)}} = 49$ $\Rightarrow a^2 r^4 = 49$ $\Rightarrow ar^2 = 7$ (since $a$ and $r$ are real).Given the sum of the first and third term is $35$:$a + ar^2 = 35$.Substituting $ar^2 = 7$ into the equation:$a + 7 = 35 \Rightarrow a = 28$.

In a geometric progression,if the ratio of the sum of the first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$,then the first term of this geometric progression is:

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