Question

If $\frac{bc}{ad}=\frac{b+c}{a+d}=3\left(\frac{b-c}{a-d}\right)$, then a, b, c, d are in

• A. A.P
• B. G.P.
• C. H.P.
• D. A.GP

Answer 1

The correct option is C H.P.

$\frac{bc}{ad}=\frac{b+c}{a+d}\Rightarrow \frac{b+c}{bc}=\frac{a+d}{bc}\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}$ ....(1)
$\frac{bc}{ad}=3\left(\frac{b-c}{a-d}\right)\Rightarrow \frac{a-d}{ad}=\frac{3(b+c)}{bc}\Rightarrow \frac{1}{d}=\frac{1}{a}+3(\frac{1}{c}-\frac{1}{b})$ ....(2)
$Let\frac{1}{b}-\frac{1}{a}=\alpha .Then\left(1\right)\Rightarrow \frac{1}{d}-\frac{1}{c}=\alpha ,$
$Substitutingfor\frac{1}{b}-\frac{1}{d}in\left(2\right),weget\frac{1}{c}+\alpha =\frac{1}{a}+\frac{3}{c}-3(\frac{1}{a}+\alpha )$
$\Rightarrow \frac{1}{c}-\frac{1}{a}+2\alpha .\therefore \frac{1}{b}=\frac{1}{a}+\alpha ,\frac{1}{c}=\frac{1}{a}+2\alpha ,\frac{1}{d}=\frac{1}{a}+3\alpha \frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in A.P. and f a, b, c, d are in H.P