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

Answer: 79

$3^{2\sin 2\theta -1},14$ and $3^{4-2\sin 2\theta }$ are in A.P.
$\Rightarrow 28=3^{2\sin 2\theta -1}+3^{4-2\sin 2\theta }$
$\Rightarrow 28=\frac{3^{2\sin 2\theta }}{3}+\frac{3^4}{3^{2\sin 2\theta }}$
$\Rightarrow 28=\frac{m}{3}+\frac{3^4}{m}\dots [$ Take $m=3^{2\sin 2\theta }]$
$\Rightarrow 84m=m^2+3^5$
$\Rightarrow m^2-84m+243=0$
$\Rightarrow (m-3)(m-81)=0$
$\Rightarrow m=3$ 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.
$\Rightarrow t_7=1+(7-1)13=79$