Question

prove that the sum of later half of 2n terms of an AP is equal to one third of the sum of the first 3n terms

Answer 1

Let the first term and common difference of the A.P. be a and d respectively.So sum of latter half 2n term , = Tn+1+Tn+2+...+T2n=(T1+T2+T3+...+Tn)+(Tn+1+Tn+2+...+T2n)−(T1+T2+T3+...+Tn)=S2n−Sn=2n2[2a+(2n−1)d]−n2[2a+(n−1)d]=n2[4a+2(2n−1)d−2a−(n−1)d]=n2[2a+(3n−1)d]=13×3n2[2a+(3n−1)d]=13×S3n=13sum of first 3n terms.
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