# If the sum of first n,2n,3n terms of an A.P. be s1,s2,s3 then prove that s3=3(s2-s1)

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by NoorVala181

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by NoorVala181

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Lets we start the question from given

n= s1

2n= s2

3n= s3

to prove= s3=3(s2-s1)

l.h.s.===s3=3n

r.h.s===3(2n-n)

=3n

so it prove that l.h.s = r.h..s

n= s1

2n= s2

3n= s3

to prove= s3=3(s2-s1)

l.h.s.===s3=3n

r.h.s===3(2n-n)

=3n

so it prove that l.h.s = r.h..s

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Sn of A.P.is given by Sn = n/2[2*a + (n-1)*d] where n= no.of terms , a= first term & d=common difference.

So S1 = Sn = n/2*[2*a + (n-1)*d] . for S2 replace '2n' for 'n'

Now S2 = S2n = n*[2*a + (2n-1)*d] for S3 replace '3n' for 'n'

Now S3 = S3n = 3n/2*[2*a + (3n-1)*d] ---------------- [1]

Hence S2 - S1 = n*[2*a + (2n-1)*d] - n/2*[2*a + (n-1)*d] simplifies to n/2*(2*a+3*n*d-d)

= n/2*[2*a +(3n-1)*d] ----- [2]

Now [1]& [2] clearly show that S3=3*(S2-S1)

So S1 = Sn = n/2*[2*a + (n-1)*d] . for S2 replace '2n' for 'n'

Now S2 = S2n = n*[2*a + (2n-1)*d] for S3 replace '3n' for 'n'

Now S3 = S3n = 3n/2*[2*a + (3n-1)*d] ---------------- [1]

Hence S2 - S1 = n*[2*a + (2n-1)*d] - n/2*[2*a + (n-1)*d] simplifies to n/2*(2*a+3*n*d-d)

= n/2*[2*a +(3n-1)*d] ----- [2]

Now [1]& [2] clearly show that S3=3*(S2-S1)