Mathematics - Sequence & Series Question with Solution | TestHub

Since, $3^{2\sin 2\alpha -1},14$ and $3^{4-2\sin 2\alpha }$ are in $\mathrm{AP}$. Therefore,

$2\times 14=3^{2\sin 2\alpha -1}+3^{4-2\sin 2\alpha }$

$\Rightarrow 28=\frac{a}{3}+\frac{3^4}{a}$, where $a=3^{2\sin 2\alpha }$

$\Rightarrow a^2-84a+243=0$

$\Rightarrow \left(a-81\right)\left(a-3\right)=0$

$\Rightarrow a=81, a=3$

$\Rightarrow 3^{2\sin 2\alpha }=3^4$ or $3^{2\sin 2\alpha }=3$

$\Rightarrow 2\sin 2\alpha =1$  $\left[\because 2\sin 2\alpha \ne 4\right]$

$\Rightarrow \sin 2\alpha =\frac{1}{2}\Rightarrow 2\alpha =30° \left[\because \sin 30°=1/2\right]$

Thus, the first three terms of the $\mathrm{AP}$ are $1,14,27$.

Hence, its fifth term $a_5=a_1+\left(5-1\right)d$

$=1+4\times 13=1+52=53$