Question

If a,b,c are in AP ,prove that (a - c)² = 4 (b² - ac).

Answer 1

Since a,b,c are in AP ,we have b = 1/2 ( a + c).

RHS = 4(b² - ac) = 4 × { 1/4 ( a + c ) ² - ac}

= (a + c)² - 4ac

= ( a - c)² = LHS

Hence

( a - c )² = 4 ( b² - ac ).

Answer 2

If a, b and c are in AP
Their common difference will be same
b - a = c - b
a + c = 2b

Square above equation on both sides
(a + c)² = (2b)²
a² + c² + 2ac = 4b²

Substract 4ac from both the sides
a² + c² + 2ac - 4ac = 4b² - 4ac
a² + c² - 2ac = 4(b² - ac)

a² + c² - 2ac is in the form of (a - c)²

∴ (a - c)² = 4(b² - ac)