Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ 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 __________.
Answer (integer)
16
Solution
Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as
<br/><br/>$$
\begin{aligned}
& \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\\\
& \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\\\
& \Rightarrow a=2
\end{aligned}
$$
<br/><br/>also sum of thes A.G.P. is $\frac{49}{2}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{2-2 d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2 d)=\frac{49}{2} \\\\
& \Rightarrow \frac{1}{4}[2-2 d+4-2 d+8+16+8 d+32+32 d]=\frac{49}{2} \\\\
& \Rightarrow 36 d+62=98
\end{aligned}
$$
<br/><br/>$\Rightarrow 36 d=36 \Rightarrow d=1$
<br/><br/>Hence, $a_4=4(a+2 d)=4(2+2 \times 1)=16$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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