Question

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

Answer 1

$(a-b)^2$, ($a^2+b^2$) and $(a + b)^2$ are in A.P., shown.

Step-by-step explanation:

To show that, $(a-b)^2$, ($a^2+b^2$) and $(a + b)^2$ are in A.P.

We know that,

If a, b, c are in AP.

b = $\dfrac{a+c}{2}$

Here, first term = $(a-b)^2$, second term = ($a^2+b^2$) and third term = $(a + b)^2$

∴ ($a^2+b^2$) = $\dfrac{(a-b)^2+(a+b)^2}{2}$

Using the algebraic identity,

$(a-b)^2+(a+b)^2=2(a^2+b^2)$

⇒ ($a^2+b^2$) = $\dfrac{2(a^2+b^2)}{2}$

⇒ ($a^2+b^2$) = ($a^2+b^2$), shown.

Thus, $(a-b)^2$, ($a^2+b^2$) and $(a + b)^2$ are in A.P., shown.