If $3^{2\sin 2\theta -1},14,3^{4-2\sin 2\theta }$ are first three terms of an A.P., then its fifth term is given by 

We have $\frac{t}{3}+\frac{81}{t}=28$ where $t=3^{2\sin 2\theta }$

$\Rightarrow t^2-84t+243=0\Rightarrow t=3,81$

But$1/9 \le t \le 9$. Thus, $t = 3.$

$\therefore a_1=1,d=13\Rightarrow a_5=a_1+4d=53$