Suppose a_1, a_2, 2, a_3, a_4 be in an arithmetico-geometric progression (with the middle …

Question

Suppose $a_1, a_2, 2, a_3, a_4$ be in an arithmetico-geometric progression (with the middle term equal to $2$). If the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\dfrac{49}{2}$, then $a_4$ is equal to \_\_\_\_\_\_\_\_.

Answer: 16

Accepted Answer

Write the five terms as $\dfrac{a-2d}{4}, \dfrac{a-d}{2}, a, 2(a+d), 4(a+2d)$ (linear part $\times$ ratio $2^k$), with middle term $a=2$.

$$\frac14+\frac12+1+6\times2+(-1+2+8)d = \frac{49}{2} \implies \frac34+7+9d=\frac{49}{2} \implies 9d = 9 \implies d=1.$$

$a_4 = 4(a+2d) = 4(4) = 16.$

Solutions · 1