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

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

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

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

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None of these

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Solution

The correct option is C ${(a b + b c + c d)}^{2}$
As $a, b, c, d$ are in $G . P .$ , therefore
$\frac{b}{a} = \frac{c}{d} = \frac{d}{c} = r$ (say)
$\Rightarrow b = a r, c = b r = a r . r = a r^{2}, d = c r = a r^{2} . r = a r^{3}$
$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a^{2} + a^{2} r^{2} + a^{2} r^{4}) (a^{2} r^{2} + a^{2} r^{4} + a^{2} r^{6}) = a^{4} r^{2} (1 + r^{2} + r^{4}) (1 + r^{2} + r^{4}) = {(a^{2} t + a^{2} r^{3} + a^{2} r^{5})}^{2} = {(a . a r + a r . a r^{2} + {a r}^{2} . a r^{3})}^{2} = {(a b + b c + c d)}^{2}$

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Similar questions

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

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

If a, b, c, d are in GP, prove that

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

a, b, c, d

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

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

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