the sum of n, 2n 3n term of an AP are 51 52 and 53 respectively
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It is given that the sum of first n ,2n and 3n terms of AP be s1, s2 and s3.
To prove:
The sum of first n terms is defined as
Taking right hand side of the given equation.
3(S_2-S_1)=\frac{3n}{2}(4a+4dn-2d-2a-dn+d)
3(S_2-S_1)=\frac{3n}{2}(2a+(3n-1)d)
L.H.S=R.H.S.
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