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 $S_5 = a(1+r+r^2+r^3+r^4) = a\frac{r^5-1}{r-1}$.The r

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$28$

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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 $S_5 = a(1+r+r^2+r^3+r^4) = a\frac{r^5-1}{r-1}$.The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.The sum of the reciprocals is $S'_5 = \frac{1}{a}(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}) = \frac{1}{a} \frac{\frac{1}{r^5}-1}{\frac{1}{r}-1} = \frac{1}{a} \frac{1-r^5}{r^4(1-r)} = \frac{r^5-1}{ar^4(r-1)}$.Given the ratio $\frac{S_5}{S'_5} = 49$:$\frac{a(r^5-1)/(r-1)}{(r^5-1)/(ar^4(r-1))} = 49$$a^2 r^4 = 49$$(ar^2)^2 = 7^2 \Rightarrow ar^2 = 7$.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$$a = 28$.

In a geometric progression,if the ratio of the sum of 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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