Q: If $ \frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3 \frac{a_2 – a_3}{a_1 – a_4} $ , then a1 , a2 , a3 , a4 are in
(A) A.P
(B) G.P
(C) H.P
(D) none of these
Ans: (C)
Sol: $\large \frac{a_1 + a_4}{a_1 a_4} = \frac{a_2 + a_3}{a_2 a_3} $
$\large \frac{1}{a_4} + \frac{1}{a_1} = \frac{1}{a_2} + \frac{1}{a_3} $
$\large \frac{1}{a_2} – \frac{1}{a_1} = \frac{1}{a_4} – \frac{1}{a_3} $
Also , $\large 3 \frac{(a_2 – a_3)}{a_2 a_3} = \frac{a_1 – a_4}{a_1 a_4}$
$\large 3(\frac{1}{a_3} – \frac{1}{a_2} ) = \frac{1}{a_4} – \frac{1}{a_1} $
So , $\large \frac{1}{a_1} , \frac{1}{a_2} , \frac{1}{a_3} , \frac{1}{a_4}$ are in A.P