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(C) Let the terms of the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given $T_2 + T_6 = \frac{70}{3}$,so $ar + ar^5 = \fra

Vedclass · Accepted Answer

$91$

Vedclass · Answer

(C) Let the terms of the geometric progression be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.Given $T_2 + T_6 = \frac{70}{3}$,so $ar + ar^5 = \frac{70}{3} \implies ar(1 + r^4) = \frac{70}{3}$.Given $T_3 \cdot T_5 = 49$,so $(ar^2)(ar^4) = 49 \implies a^2r^6 = 49 \implies ar^3 = 7$ (since terms are positive).Thus,$a = \frac{7}{r^3}$.Substituting $a$ into the first equation: $\frac{7}{r^3} \cdot r(1 + r^4) = \frac{70}{3} \implies \frac{7}{r^2}(1 + r^4) = \frac{70}{3} \implies \frac{1 + r^4}{r^2} = \frac{10}{3}$.Let $r^2 = t$. Then $\frac{1 + t^2}{t} = \frac{10}{3} \implies 3t^2 - 10t + 3 = 0$.Solving for $t$: $(3t - 1)(t - 3) = 0$,so $t = 3$ or $t = \frac{1}{3}$.Since the $G$.$P$. is increasing,$r > 1$,so $r^2 = 3$.We need to find $T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4)$.Substituting $ar^3 = 7$ and $r^2 = 3$: $7(1 + 3 + 3^2) = 7(1 + 3 + 9) = 7(13) = 91$.

In an increasing geometric progression of positive terms,the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is:

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