Question

If $3^{2 \sin 2 \theta-1}, 14$ and $3^{4-2 \sin 2 \theta}$ are the first three terms of an A.P., then find its $7^{\text {th }}$ term.

Answer 1

$3^{2 \sin 2 \theta-1}, 14$ and $3^{4-2 \sin 2 \theta}$ are in A.P.
$\Rightarrow 28=3^{2 \sin 2 \theta-1}+3^{4-2 \sin 2 \theta} $
$\Rightarrow 28=\frac{3^{2 \sin 2 \theta}}{3}+\frac{3^{4}}{3^{2 \sin 2 \theta}}$
$\Rightarrow 28=\frac{m}{3}+\frac{3^{4}}{m}\ldots\left[\right.$ Take $\left.m=3^{2 \sin 2 \theta}\right]$
$\Rightarrow 84 m=m^{2}+3^{5}$
$\Rightarrow m^{2}-84 m+243=0$
$\Rightarrow(m-3)(m-81)=0$
$\Rightarrow m=3$ or $m=81=3^{4}$
$\Rightarrow 3^{2 \sin 2 \theta}=3 $
or $ 3^{2 \sin 2 \theta}=3^{4}$
But, $2 \sin 2 \theta=4$ is not possible.
$\Rightarrow 2 \sin 2 \theta=1$
$\Rightarrow 2 \theta=30^{\circ}$
$\Rightarrow 3^{\circ}, 14$ and $3^{3}$ are in A.P.
i.e. $1,14$ and $27$ are in A.P.
$\Rightarrow t_{7}=1+(7-1) 13=79$