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

Open in App

Solution

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

Suggest Corrections

Similar questions

a, b, c

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

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

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

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

△ A B C

a^{2}, b^{2}, c^{2}

A . P

cot A, cot B, cot C

A . P

Join BYJU'S Learning Program