Question

show that (a - b)^2 whole square, (a^2 + b^2) and (a+b)^2 are in AP.​

Answer 1

let a,b,c are in ap then b-a=c-b

Step-by-step explanation:

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

so.

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

similarly

(a+b)^2=a^2+b^+2ab

so

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

then we conclude that (a-b)^2,(a^2+b^2),(a+b)^2 are in ap