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

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Solution

Taking LHS
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$
putting values $b = a r, c = a r^{2}, d = a r^{3}$
$(a^{2} + (a r)^{2} + (a r^{2})^{2}) ((a r)^{2} + (a r^{2})^{2} + (a r^{3})^{2})$
$= a^{2} (1 + r^{2} + r^{4}) a^{2} (r^{2} + r^{4} + r^{6})$
$= a^{4} r^{2} (1 + r^{2} + r^{4})^{2}$
RHS $= (a b + b c + c d)^{2}$
$= (a \times a r + a r \times a r^{2} + a r^{2} \times a r^{3})^{2}$
$= (a^{2} r + a^{2} r^{3} + a^{2} r^{5})^{2}$
$= a r^{2} (1 + r^{2} + r^{4})^{2}$
$=$ LHS.

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