show that (a-b)²,(a²+b²) & (a+b)²are in A.P.
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=(a-b)²
second term=
d=second term - first term
=
=2ab
third term= first term+2(d)
=
Hence , it forms an A.P. :
(a-b)²,(a²+b²) & (a+b)²
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.
:)