Suppose $a_1\mathit{,}{a}_2\mathit{,}2\mathit{,}{a}_3\mathit{,}{a}_4$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\frac{49}{2}$, then $a_4$ is equal to ______________

Given,

$a_1\mathit{,}{a}_2\mathit{,}2\mathit{,}{a}_3\mathit{,}{a}_4$ be in an arithmetico-geometric progression,

And the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\frac{49}{2}$

So,common ratio Here $r = 2$ and let common difference $= d$

And we know that, 

$a_2=\left(a_1+d\right)\times 2$

$2=\left(a_1+2d\right)\times 2^2 ..........\left(i\right)$

$a_3=\left(a_1+3d\right)\times 2^3$

$a_4=\left(a_1+4d\right)\times 2^4$

Now adding all above equations we get,

$31a_1+98d=\frac{49}{2} .........\left(ii\right)$

Now from $\left(i\right) \& \left(ii\right)$

$a_1=0,d=\frac{1}{4}$

Hence, $a_4=16(0+1)=16$