If 3^{2 sin 2 alpha - 1},14,and 3^{4 - 2 sin | vedclass

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(A) Let the terms be $a_1 = 3^{2 \sin 2 \alpha - 1}$,$a_2 = 14$,and $a_3 = 3^{4 - 2 \sin 2 \alpha}$.Since they are in $A.P.$,$2a_2 = a_1 + a_3$.$2(14)

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(A) Let the terms be $a_1 = 3^{2 \sin 2 \alpha - 1}$,$a_2 = 14$,and $a_3 = 3^{4 - 2 \sin 2 \alpha}$.Since they are in $A.P.$,$2a_2 = a_1 + a_3$.$2(14) = 3^{2 \sin 2 \alpha - 1} + 3^{4 - 2 \sin 2 \alpha} = 28$.Let $x = 3^{2 \sin 2 \alpha}$. Then the equation becomes $\frac{x}{3} + \frac{81}{x} = 28$.$x^2 - 84x + 243 = 0$.$(x - 81)(x - 3) = 0$,so $x = 81$ or $x = 3$.If $x = 3$,$3^{2 \sin 2 \alpha} = 3^1 \implies 2 \sin 2 \alpha = 1 \implies \sin 2 \alpha = 0.5$.Then $a_1 = 3^{1-1} = 1$ and $a_2 = 14$. The common difference $d = 14 - 1 = 13$.The $6^{th}$ term $T_6 = a_1 + 5d = 1 + 5(13) = 1 + 65 = 66$.If $x = 81$,$3^{2 \sin 2 \alpha} = 3^4 \implies 2 \sin 2 \alpha = 4 \implies \sin 2 \alpha = 2$,which is impossible.

If $3^{2 \sin 2 \alpha - 1}$,$14$,and $3^{4 - 2 \sin 2 \alpha}$ are the first three terms of an $A.P.$ for some $\alpha$,then the sixth term of this $A.P.$ is

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