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  4. If (a2 a3/a1 a4)=(a2+a3/a1+a4)=(3(a2-a3)/a1-a4), then a1, a2, a3, a4 are in

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Q. 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

Sequences and Series

Solution:

$ \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