Question
If a, b, c, are in G.P., prove that: $\frac{1}{\text{a}^2+\text{b}^2},\frac{1}{\text{b}^2+\text{c}^2},\frac{1}{\text{c}^2+\text{d}^2}\text{ are in G.P.}$

Answer

a, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ $\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{b}^2}\Big)^2+\frac{2}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{c}^2}\Big)^2$ $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{ac}}\Big)^2+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{bd}}\Big)^2$ [Using (1)] $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2\text{c}^2}+\frac{1}{\text{a}^2\text{d}^2}+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{d}^2}$ [Using (1)]$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)+\frac{1}{\text{b}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{a}^2+\text{b}^2}\Big)\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\therefore\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big),\Big(\frac{1}{\text{c}^2+\text{d}^2}\Big)\text{ and are also in G.P.}$

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