Question

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

Answer 1

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

$\therefore$ $RHS=4(b^2-ac)=4\times \left\{\frac{1}{4}\right(a+c{)}^2-ac\}$

$=(a+c{)}^2-4ac=(a-c{)}^2=LHS$

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