Question

If $\frac{a_2{a}_3}{a_1{a}_4}=\frac{a_2+a_3}{a_1+a_4}=\frac{3\left(a_2-a_3\right)}{a_1-a_4}$, then $a_1,a_2,a_3,a_4$ are in

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

Answer: (C)

$\frac{a_1+a_4}{a_1+a_4}=\frac{a_2+a_3}{a_2{a}_3}$
$\Rightarrow \frac{1}{a_1}+\frac{1}{a_4}=\frac{1}{a_2}+\frac{1}{a_3}$........(1)
$\text{ also }\frac{a_1-a_4}{a_1{a}_4}=\frac{3\left(a_2-a_3\right)}{a_2{a}_3}$
$\Rightarrow \frac{1}{a_4}-\frac{1}{a_1}=3\left(\frac{1}{a_3}-\frac{1}{a_2}\right)$.......(2)
by (1) & (2) $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\frac{1}{a_4}$
Hence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\frac{1}{a_4}$ are in $AP$
$\Rightarrow a_1,a_2,a_3,a_4$ are in HP