Show that (a - b) 2 , (a 2 + b 2 ) and (a + b) 2 are in AP.

Question

Philoid · Accepted Answer

Consider (a² + b²) - (a - b)²

= (a² + b²) - (a² + b² - 2ab)

= 2ab

Consider (a + b)² - (a² + b²)

= (a² + b² + 2ab) - (a² + b²)

= 2ab

Since, the difference between consecutive terms is constant, therefore the terms are in AP.

Show that (a - b)2, (a2 + b2) and (a + b)2 are in AP.