If a, b, c are in AP, then $(a c)^{2}$ equals.

{(2 b^{2} - \frac{(a^{2} + c^{2})}{2})}^{2}

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4 (b^{2} + a c)

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4 b^{2} a c

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b^{2} 4 a c

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Solution

The correct option is C ${(2 b^{2} - \frac{(a^{2} + c^{2})}{2})}^{2}$
If $a, b, c$ are in A.P. then,

$2 b = a + c$

squaring both side,

$\Rightarrow 4 b^{2} = a^{2} + c^{2} + 2 a c$

$\Rightarrow 2 a c = 4 b^{2} - a^{2} - c^{2}$

$(a c)^{2} = {(2 b^{2} - \frac{(a^{2} + c^{2})}{2})}^{2}$

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(a - c)^{2} = 4 (b^{2} - a c)

If a, b, c are in AP, prove that $(a - c)^{2} = 4 (b^{2} - a c)$ .

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(a - c)^{2} = 4 (b^{2} - a c)

a, b, c

(a - c)^{2} = 4 (b^{2} - a c)

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