If sum of (n+1) terms of an arithmetic sequence is pn^2+qn+r, prove that p+r=q
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Given Sn=Pn+Qn2
∴Tn=Sn−Sn−1=Pn+Qn2−[P(n−1)+Q(n−1)2]=P−Q+2nQ
Thus common difference, d=Tn−Tn−1=P−Q+2nQ−[P−Q+2(n−1)Q]=2Q
Hence, option 'A' is correct.
Given Sn=Pn+Qn2
∴Tn=Sn−Sn−1=Pn+Qn2−[P(n−1)+Q(n−1)2]=P−Q+2nQ
Thus common difference, d=Tn−Tn−1=P−Q+2nQ−[P−Q+2(n−1)Q]=2Q
Hence, option 'A' is correct.
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