Question

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

Answer 1

The given numbers are (a – b)², (a² + b²) and (a² + b²)

Now, 

(a² + b²) – (a – b)² 

= a² + b² – (a² – 2ab + b²) 

= a² + b² – a² + 2ab – b²

= 2ab

(a + b)² – (a² + b²) 

= a² + 2ab + b² – a² – b²

= 2ab

So, (a² + b²) – (a – b)² 

= (a + b)² – (a² + b²) 

= 2ab   ...(Constant)

Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.