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

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Solution

Since a, b, c are in AP, we have $b = \frac{1}{2} (a + c)$

$∴$ $R H S = 4 (b^{2} - a c) = 4 \times {\frac{1}{4} (a + c)^{2} - a c}$

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

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

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