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

1.	Show that (a–b)2, (a2+b2) and (a + b)2 are in A.P.
Answer» The terms given below are : (a–b)², (a²+b²) and (a + b)² Common difference, d₁ = a² + b² – (a – b)² d₁ = a² + b² – (a² + b² - 2ab) d₁ = a² + b² – a² - b² + 2ab d₁= 2ab Common difference, d₂ = (a + b)² – (a² + b²) d₂ = a² + b² + 2ab – a² –b² d₂ = 2ab Since, d₁ = d₂ i.e. the common difference is same. Therefore, the given terms are in A.P.

Answer»

The terms given below are :

(a–b)², (a²+b²) and (a + b)²

Common difference, d₁ = a² + b² – (a – b)²

d₁ = a² + b² – (a² + b² - 2ab)

d₁ = a² + b² – a² - b² + 2ab

d₁= 2ab

Common difference, d₂ = (a + b)² – (a² + b²)

d₂ = a² + b² + 2ab – a² –b²

d₂ = 2ab

Since, d₁ = d₂ i.e. the common difference is same.

Therefore, the given terms are in A.P.

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