In an increasing geometric series,the sum of | vedclass

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(C) Let the geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 1$ for an increasing series.Given $T_2 + T_6 = \frac{25}{2} \Rightarrow ar

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$35$

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(C) Let the geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 1$ for an increasing series.Given $T_2 + T_6 = \frac{25}{2} \Rightarrow ar(1 + r^4) = \frac{25}{2} \dots (1)$Given $T_3 \cdot T_5 = 25$ $\Rightarrow (ar^2)(ar^4) = 25$ $\Rightarrow a^2r^6 = 25$ $\Rightarrow ar^3 = 5$ (since $a, r > 0$)From $(1)$,$ar + ar^5 = \frac{25}{2}$. Substituting $a = \frac{5}{r^3}$:$\frac{5}{r^3} \cdot r + \frac{5}{r^3} \cdot r^5 = \frac{25}{2} \Rightarrow \frac{5}{r^2} + 5r^2 = \frac{25}{2}$Dividing by $5$: $\frac{1}{r^2} + r^2 = \frac{5}{2} \Rightarrow 2r^4 - 5r^2 + 2 = 0$$(2r^2 - 1)(r^2 - 2) = 0 \Rightarrow r^2 = 2$ (since $r > 1$)Then $a = \frac{5}{r^3} = \frac{5}{2\sqrt{2}}$.The sum of $4^{	ext{th}}, 6^{	ext{th}}, 8^{	ext{th}}$ terms is $ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4)$.$= 5(1 + 2 + 4) = 5(7) = 35$.

In an increasing geometric series,the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25$. Then,the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is equal to

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