Answers

2015-03-12T17:06:38+05:30
A,b,c...
common difference=d
a+d=b
a+2d=c
sum=a+b+c
=a+a+d+a+2d=3a+3d

a+c=a+a+2d=2a+2d
b+c-2a=a+d+a+2d-2a=2a+3d-2a=3d
2(b-a)=2(a+d-a)=2d
\frac{(a+c)(b+c-2a) }{2(b-a)}  \\  \\ 
=  \frac{(2a+2d)3d}{2d}  \\  \\ 
= \frac{(2a+2d)3}{2}  \\  \\ 
=\frac{6a+6d }{2} =3a+3d

hence proved...

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