The sum of the 3^{rd} and the 4^{th} terms of | vedclass

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(D) Let the first three terms of the $G.P.$ be $a, ar, ar^2$.According to the problem,the product of the first three terms is $1000$:$a(ar)(ar^2) = 10

Vedclass · Accepted Answer

$320$

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(D) Let the first three terms of the $G.P.$ be $a, ar, ar^2$.According to the problem,the product of the first three terms is $1000$:$a(ar)(ar^2) = 1000$ $\Rightarrow (ar)^3 = 1000$ $\Rightarrow ar = 10$.The sum of the $3^{rd}$ and $4^{th}$ terms is $60$:$ar^2 + ar^3 = 60 \Rightarrow ar(r + r^2) = 60$.Substituting $ar = 10$ into the equation:$10(r + r^2) = 60 \Rightarrow r^2 + r - 6 = 0$.Factoring the quadratic equation:$(r + 3)(r - 2) = 0 \Rightarrow r = 2$ or $r = -3$.Case $1$: If $r = 2$,then $a(2) = 10 \Rightarrow a = 5$.Case $2$: If $r = -3$,then $a(-3) = 10 \Rightarrow a = -10/3$.Since the first term $a$ must be positive,we choose $a = 5$ and $r = 2$.The $7^{th}$ term is given by $T_7 = ar^6 = 5(2)^6 = 5 	imes 64 = 320$.

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive,then its $7^{th}$ term is

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