If $a, b, c, d$ are in $G . P$ then prove that $a^{2} + b^{2}, b^{2} + c^{2}, c^{2} + d^{2}$ are also in $G . P$

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Solution

a, b, c, d are in G.P
$a^{2} + b^{2} = a^{2} (1 + r^{2})$
$b^{2} + c^{2} = b^{2} (1 + r^{2})$
$c^{2} + d^{2} = c^{2} (1 + r^{2})$
$R_{1} = \frac{b^{2} + c^{2}}{a^{2} + b^{2}} = \frac{b^{2}}{a^{2}} = r^{2}$
$R_{2} = \frac{c^{2} + d^{2}}{b^{2} + c^{2}} = \frac{c^{2}}{b^{2}} = r^{2}$
$R_{1} = R_{2} = r^{2} = R$
So we have
$a^{2} + b^{2}$ ;
$b^{2} + c^{2} = R (a^{2} + b^{2})$
$c^{2} + d^{2} = R (b^{2} + c^{2})$
are in G.P.

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