If $3^{2\sin 2\alpha -1},14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{4-2\sin 2\alpha }$ are

the first three terms of an A.P for some $\alpha$ ,  then the sixth term of this A.P. is:

let $3^{2\sin 2\alpha }=t,t\in \left[\frac{1}{9},9\right]$

Hence, $\frac{t}{3},14,\frac{81}{t}$ terms are in arithmetic progression. 

If $a,b,c$ are in arithmetic progression then $2b=a+c$

it implies that $28=\frac{t}{3}+\frac{81}{t}\Rightarrow t^2-84t+243=0\Rightarrow t=3,81$

suppose $t=3$, so that $3^{2\sin 2\alpha }=3\Rightarrow \sin 2\alpha =\frac{1}{2}$

when $t=3$ the terms are $1,14,27,...$

First term is 1, common difference is 13

$a_6=a+\left(5\right)d=1+5\left(13\right)=66$