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(B) Let the first term be $a > 0$ and the common ratio be $r > 0$.Given the sum of the second,third,and fourth terms is $3$:$ar + ar^2 + ar^3 = 3$  --

Vedclass · Accepted Answer

$\frac{1}{26}(3^{50}-1)$

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(B) Let the first term be $a > 0$ and the common ratio be $r > 0$.Given the sum of the second,third,and fourth terms is $3$:$ar + ar^2 + ar^3 = 3$  --- $(1)$Given the sum of the sixth,seventh,and eighth terms is $243$:$ar^5 + ar^6 + ar^7 = 243$$r^4(ar + ar^2 + ar^3) = 243$Substituting $(1)$ into this equation:$r^4(3) = 243$ $\Rightarrow r^4 = 81$ $\Rightarrow r = 3$ (since $r > 0$).Substituting $r = 3$ into $(1)$:$a(3) + a(9) + a(27) = 3$$39a = 3 \Rightarrow a = \frac{3}{39} = \frac{1}{13}$.The sum of the first $50$ terms $S_{50}$ is given by:$S_{50} = \frac{a(r^{50}-1)}{r-1} = \frac{\frac{1}{13}(3^{50}-1)}{3-1} = \frac{1}{26}(3^{50}-1)$.

If the sum of the second,third and fourth terms of a positive term $G.P.$ is $3$ and the sum of its sixth,seventh and eighth terms is $243,$ then the sum of the first $50$ terms of this $G.P.$ is

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