Let a_1, a_2, a_3, ldots be a G.P. of increas | vedclass

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(D) Let the $G.P.$ be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $a_6 + a_8 = 2 \implies ar^5 + ar^7 = 2 \implies ar^5(1 + r^2) = 2$.Given

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$3$

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(D) Let the $G.P.$ be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $a_6 + a_8 = 2 \implies ar^5 + ar^7 = 2 \implies ar^5(1 + r^2) = 2$.Given $a_3 	imes a_5 = \frac{1}{9} \implies (ar^2)(ar^4) = \frac{1}{9} \implies a^2r^6 = \frac{1}{9} \implies ar^3 = \frac{1}{3}$ (since $a, r > 0$).Divide the first equation by the second: $\frac{ar^5(1 + r^2)}{ar^3} = \frac{2}{1/3} \implies r^2(1 + r^2) = 6 \implies r^4 + r^2 - 6 = 0$.Let $x = r^2$,then $x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0$. Since $r^2 > 0$,$r^2 = 2$.Then $a(2)^{3/2} = \frac{1}{3} \implies a = \frac{1}{3 	imes 2\sqrt{2}} = \frac{1}{6\sqrt{2}}$.We need to evaluate $6(a_2 + a_4)(a_4 + a_6) = 6(ar + ar^3)(ar^3 + ar^5) = 6(ar(1 + r^2))(ar^3(1 + r^2)) = 6a^2r^4(1 + r^2)^2$.Substitute $a^2r^6 = \frac{1}{9} \implies a^2r^4 = \frac{1}{9r^2} = \frac{1}{18}$.Expression $= 6 	imes \frac{1}{18} 	imes (1 + 2)^2 = \frac{1}{3} 	imes 9 = 3$.

Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text{th}}$ and $8^{\text{th}}$ terms be $2$ and the product of its $3^{\text{rd}}$ and $5^{\text{th}}$ terms be $\frac{1}{9}$. Then $6(a_2 + a_4)(a_4 + a_6)$ is equal to

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