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

$320$

Vedclass · Answer

(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 $1000$:$a(ar)(ar^2) = 1000 \Rightarrow (ar)^3 = 1000 \Rightarrow ar = 10$.Thus,$a = \frac{10}{r}$.The sum of the $3^{rd}$ term $(ar^2)$ and the $4^{th}$ term $(ar^3)$ is $60$:$ar^2 + ar^3 = 60 \Rightarrow ar^2(1 + r) = 60$.Substituting $ar = 10$:$10r(1 + r) = 60 \Rightarrow r(1 + r) = 6 \Rightarrow r^2 + r - 6 = 0$.Solving the quadratic equation: $(r + 3)(r - 2) = 0$,so $r = 2$ or $r = -3$.If $r = 2$,then $a = \frac{10}{2} = 5$ (positive).If $r = -3$,then $a = \frac{10}{-3} = -\frac{10}{3}$ (negative,rejected).Using $a = 5$ and $r = 2$,the $7^{th}$ term is $T_7 = ar^6 = 5(2^6) = 5 	imes 64 = 320$.

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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