Question
If $a, b, c$ and $d$ are in proportion, prove that: $ a b c d\left[\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right]= a ^2+ b ^2+ c ^2+ d ^2\right. $
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Answer
$\begin{aligned} & \because a _{,} b , c _r d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k (\text { say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=a b c d\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right) \\ & =b k \cdot b \cdot d k \cdot d\left[\frac{1}{b^2 k^2}+\frac{1}{b^2}+\frac{1}{d^2 k^2}+\frac{1}{d^2}\right] \\ & =k^2 b^2 d^2\left[\frac{d^2+d^2 k^2+b^2+b^2 k^2}{b^2 d^2 k^2}\right] \\ & = d ^2\left(1+ k ^2\right)+ b ^2(1+ k 2) \\ & =\left(1+ k ^2\right)\left( b ^2+ d ^2\right) \\ & \text { R.H.S. }= a ^2+ b ^2+ c + d ^2 \\ & = b ^2 k ^2+ b ^2+ d ^2 k ^2+ d ^2 \\ & = b ^2\left( k ^2+1\right)+ d ^2\left( k ^2+1\right) \\ & =\left( k ^2+1\right)\left( b ^2+ d ^2\right) \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
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