Question

terms of an AP.
[CBSE 2009
5. Show that (a - b)2, (a2 + b2) and (a + b)2 are in AP.
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Answer 1

Answer:

please mark as brainliest please

Step-by-step explanation:

a2 - a1 = ( a² + b² ) - ( a - b )²

= a² + b² - a² + 2ab - b²

= 2ab -----( 1 )

a3 - a2 =( a + b )² - ( a² + b² )

= a² + 2ab + b² - a² - b²

= 2ab ------( 2 )

From ( 1 ) and ( 2 ) ,

a2 - a1 = a3 - a1 = 2ab = common difference

Therefore ,

Above threeterms are in A.P

I hope this helps you

Answer 2

(a-b)^2,(a^2+b^2),(a+b)^2.....

If Cd is same they r in AP

a^2+b^2-a^2-b^2+2ab=d

2ab

a^2+b^2+2ab-a^2-b^2=d

2ab

as Cd is same it is in AP