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

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Solution

To prove: $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d^{2})$

$\Rightarrow (b^{2} + c^{2})^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$

we have, $a, b = a r, c = a r^{2}, d = a r^{3}$

$(a^{2} + b^{2}) (c^{2} + d^{2})$

$= (a^{2} + a^{2} r^{2}) (a r^{4} + a r^{6})$

$= a^{2} (1 + r^{2}) \times a^{2} r^{4} (1 + r^{2})$

$= a^{4} r^{4} (1 + r^{2})^{2}$

$= (a^{2} r^{2} + a^{2} r^{4})^{2}$

$= (b^{2} + c^{2})^{2}$

LHS=RHS

Hence proved.

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