Question

show that (a-b)²,(a²+b²) & (a+b)²are in A.P.

Answer 1

Let first term be :
$(a - b)(a - b)$
=(a-b)²
second term=
${a }^{ 2} + {b}^{2}$
d=second term - first term
=
${a}^{2} + {b}^{2} - ( {a}^{2} - 2ab + {b}^{2} )$
=2ab
third term= first term+2(d)
=
${a}^{2} - 2ab + {b }^{2} + 2(2ab)$

$= {a}^{2} + 2ab + {b}^{2}$
$= {(a + b)}^{2}$
Hence , it forms an A.P. :
(a-b)²,(a²+b²) & (a+b)²

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 threeterms are in A.P

I hope this helps you.

:)