Question

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

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Answer 1

Answer:

(a−b)² ,(a²+b²) and (a+b)² are in AP.

So, difference between two consecutive terms will be same.

(a² +b²) −(a−b) 2 =(a+b)² −(a² +b²)(a²+b²)−(a² +b²−2ab)=a² +b² +2ab−a² −b²

2ab=2ab.

Hence given terms are in AP.

Explanation:

Hope it helps❣️☺️

Answer 2

Hi ,

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 three terms are in A.P

hope it helps you ❤️

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