Suppose a_1, a_2, 2, a_3, a_4

Question

Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$

Accepted Answer

Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as

$\begin{gathered} \frac{(a-2d)}{4},\frac{(c-d)}{2},a,2(a+d),4(a+2d) \ {a}_{1},{a}_{2},2,{a}_{3},{a}_{4} \ \Rightarrow a=2 \end{gathered}$

also sum of thes A.G.P. is $\frac{49}{2}$

$\begin{gathered} \Rightarrow \frac{2-2d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2d)=\frac{49}{2} \ \Rightarrow \frac{1}{4}[2-2d+4-2d+8+16+8d+32+32d]=\frac{49}{2} \ \Rightarrow 36d+62=98 \end{gathered}$

$\Rightarrow 36d=36\Rightarrow d=1$

Hence, ${a}_{4}=4(a+2d)=4(2+2	imes 1)=16$

Solutions · 1