If ${(x - a + b)}^{2} + {(x - b + c)}^{2} = 0$ , then $a, b, c$ are in

A.P.

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G.P.

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H.P.

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A.G.P.

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Solution

The correct option is A A.P.
Assuming $x, a, b$ and $c$ are real numbers

Since sum of two square number is zero, so both numbers have to be zero for the given condition
$\Rightarrow$ $x - a + b$ $=$ 0, and $x - b + c$ $=$ 0
$i . e$ $x$ $=$ $a - b$ $=$ $b - c$

$\Rightarrow$ $2 b$ $=$ $a + c$

so $a, b, c$ are in A.P

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