Suppose a_1, a_2, 2, a_3, a_4 are in an arith | vedclass

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(C) Let the terms of the arithmetico-geometric progression be of the form $(a+nd)r^n$. Given the $3^{rd}$ term is $2$ and the common ratio $r=2$,we ca

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$16$

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(C) Let the terms of the arithmetico-geometric progression be of the form $(a+nd)r^n$. Given the $3^{rd}$ term is $2$ and the common ratio $r=2$,we can write the terms as: $a_1 = \frac{a-2d}{4}, a_2 = \frac{a-d}{2}, a_3 = a, a_4 = 2(a+d), a_5 = 4(a+2d)$. Since the $3^{rd}$ term is $2$,we have $a=2$. The terms are: $\frac{2-2d}{4}, \frac{2-d}{2}, 2, 2(2+d), 4(2+2d)$. Sum of the $5$ terms is: $\frac{2-2d}{4} + \frac{2-d}{2} + 2 + 4+2d + 8+8d = \frac{49}{2}$. $\frac{1-d}{2} + 1 - \frac{d}{2} + 14 + 10d = \frac{49}{2}$. $1 - d + 14 + 10d = \frac{49}{2}$. $15 + 9d = 24.5$. $9d = 9.5$. Wait,re-evaluating: The terms are $(a)r^0, (a+d)r^1, (a+2d)r^2, (a+3d)r^3, (a+4d)r^4$. Given $a_3 = (a+2d)r^2 = 2$. With $r=2$,$(a+2d)4 = 2 \Rightarrow a+2d = 0.5$. Sum $S_5 = a + (a+d)2 + (a+2d)4 + (a+3d)8 + (a+4d)16 = 49/2$. $a + 2a + 2d + 4a + 8d + 8a + 24d + 16a + 64d = 24.5$. $31a + 98d = 24.5$. Solving $a+2d=0.5$ and $31a+98d=24.5$: $a = 0.5 - 2d$. $31(0.5 - 2d) + 98d = 24.5$ $\Rightarrow 15.5 - 62d + 98d = 24.5$ $\Rightarrow 36d = 9$ $\Rightarrow d = 0.25$. $a = 0.5 - 2(0.25) = 0$. $a_4 = (a+3d)r^3 = (0 + 3(0.25)) 	imes 2^3 = 0.75 	imes 8 = 6$. Re-checking the provided solution logic: The terms are $\frac{a-2d}{4}, \frac{a-d}{2}, a, 2(a+d), 4(a+2d)$. Sum: $\frac{a}{4} - \frac{d}{2} + \frac{a}{2} - \frac{d}{2} + a + 2a + 2d + 4a + 8d = \frac{49}{2}$. $(0.25+0.5+1+2+4)a + (-0.5-0.5+2+8)d = 24.5$. $7.75a + 9d = 24.5$. Given $a=2$,$7.75(2) + 9d = 24.5$ $\Rightarrow 15.5 + 9d = 24.5$ $\Rightarrow 9d = 9$ $\Rightarrow d=1$. $a_4 = 2(a+d) = 2(2+1) = 6$. There is a discrepancy in the provided solution's final step. Based on the sequence $a_1, a_2, 2, a_3, a_4$,$a_4$ is the $5^{th}$ term. $a_5 = 4(a+2d) = 4(2+2) = 16$.

Suppose $a_1, a_2, 2, a_3, a_4$ are 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 $...........$.

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