Question

Show that (a – b)², (a² + b²) and (a + b)² are in A.P.

Answer 1

Here, we are given three terms and we need to show that they are in A.P.,

First term (a₁) = (a – b)²

Second term (a₂) = (a² + b²)

Third term (a₃) = (a + b)²

So in an A.P., the difference of two adjacent terms is always constant. So to prove that terms are in A.P. we find the common difference, we get

d = a₂ – a₁

d = (a² + b²) – (a – b)²

d = (a² + b²) – (a² + b² – 2ab)

d = a² + b² – a² – b² + 2ab

d = 2ab   ...(1)

Also

d = a₃ – a₂

d = (a + b)² – (a² + b²)

d = a² + b² + 2ab – a² – b²

d = 2ab   ...(2)

Now since in equations (1) and (2) the value of d are equal we can say that these term are in A.P. withn 2ab ast the commnon difference.

Hence proved.