Prove the following
Question:

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:

  1. (1) 66

  2. (2) 81

  3. (3) 65

  4. (4) 78


Correct Option: 1

Solution:

Given that $3^{2 \sin 2 \alpha-1}, 14,3^{4-2 \sin 2 \alpha}$ are in A.P.

So, $3^{2 \sin 2 \alpha-1}+3^{4-2 \sin 2 \alpha}=28$

$\Rightarrow \frac{3^{2 \sin 2 \alpha}}{3}+\frac{81}{3^{2 \sin 2 \alpha}}=28$

Let $3^{2 \sin 2 \alpha}=x$

$\Rightarrow \frac{x}{3}+\frac{81}{x}=28$

$\Rightarrow x^{2}-84 x+243=0 \Rightarrow x=81, x=3$

When $x=81 \Rightarrow \sin 2 \alpha=2$ (Not possible)

When $x=3 \Rightarrow \alpha=\frac{\pi}{12}$

$\therefore a=3^{0}=1, d=14-1=13$

$a_{6}=a+5 d=1+65=66$

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