Answers

2016-02-21T21:34:35+05:30
s1=n/2[2a+(n-1)d]
s2=2n/2[2a+(2n-1)d]
s3=3n/2[2a+(3n-1)d]
therefore,
s3=3(s2-s1)
    =3{2n/2[2a+(2n-1)d]-3n/2[2a+(3n-1)d]}
=3n/2{4a+2(2n-1)d-2a-(n-1)d}
=3n/2{2a+(4n-n+1-2)}
=3n/2[2a+(3n-1)d]=s3

hope ur query gets solved by my ans......
cheers..................
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