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

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Solution

Given $a, b, c$ are in A.P. then $b - a = c - b$ or, $b = \frac{a + c}{2}$ ......(1).
Now
$(b^{2} - a c)$
$= \frac{(a + c)^{2}}{4} - a c$ [ Using (1)]
$= \frac{(a - c)^{2}}{4}$ .
or, $4 (b^{2} - a c) = (a - c)^{2}$ .

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