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

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Solution

RHS $= 4 b^{2} - 4 a c$
$= 4 {(\frac{a + c}{2})}^{2} - 4 a c$
$= \frac{4}{4} (a^{2} + c^{2} + 2 a c) - 4 a c$
$= a^{2} + c^{2} - 2 a c$
$= (a - c)^{2}$
$= L H S$

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