Suppose a _ 1 , a _ 2 , 2, a _ 3 , a _ 4 be in an...

Given,

${a}_{1}, {a}_{2}, 2, {a}_{3}, {a}_{4}$ 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 $\dfrac{49}{2}$

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,

$31{a}_{1} + 98d = \dfrac{49}{2} . . . . . . . . . \left(ii\right)$

Now from \left(i\right) \text{ and } \left(ii\right)

${a}_{1} = 0, d =$\dfrac{1}{4}

Hence, ${a}_{4} = 16 ( 0 + 1 ) = 16$