The numbers 3 ^ 2 2α - 1 , 14 and 3 ^ 4 - 2 2α form first three...

Since, ${3}^{2\sin 2\alpha - 1}, 14$ and ${3}^{4 - 2\sin 2\alpha} are in$AP$. Therefore,</p><p>$2 \times 14 = {3}^{2\sin 2\alpha - 1} + {3}^{4 - 2\sin 2\alpha}$</p><p>$⇒ 28 = \dfrac{a}{3} + \dfrac{34}{a}$, where$a = {3}^{2\sin 2\alpha}$</p><p>$⇒ a^{2} - 84a + 243 = 0$</p><p>$⇒ \left(a - 81\right)\left(a - 3\right) = 0$</p><p>$⇒ a = 81, a = 3$</p><p>$⇒ {3}^{2\sin 2\alpha} = {3}^{4}$or${3}^{2\sin 2\alpha} = 3$</p><p>$⇒ 2\sin 2\alpha = 1 \left[∵ 2\sin 2\alpha \ne4\right]$</p><p>$⇒ \sin 2\alpha = \dfrac{1}{2}$⇒ 2\alpha =30 ° \left[∵ \sin 30 ° = 1 / 2\right]$

Thus, the first three terms of the $AP$ are $1, 14, 27$.

Hence, its fifth term ${a}_{5} = {a}_{1} + \left(5 - 1\right) d$

$= 1 + 4 \times13 = 1 + 52 = 53$