If Sn denotes the sum of first n terms of an A.P. and S3n−Sn−1S2n−S2n−1=31, then the value of n is
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Solution
Given : S3n−Sn−1S2n−S2n−1=31 ⇒3n2[2a+(3n−1)d]−n−12[2a+(n−2)d]T2n=31(∵S2n−S2n−1=T2n)⇒2a(3n−(n−1))+d(3n(3n−1)−(n−1)(n−2))2T2n=31⇒2a(2n+1)+d(8n2−2)2T2n=31⇒2(2n+1)[a+(2n−1)d]2T2n=31⇒(2n+1)T2nT2n=31⇒2n+1=31∴n=15