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Question Number 55859 by Easyman32 last updated on 05/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

S_n =(n/2)[2a+(n−1)d] S_(n−1) =((n−1)/2)[2a+(n−2)d] S_(n−2) =((n−2)/2)[2a+(n−3)d] d=(n/2)[2a+(n−1)d]+((n−2)/2)[2a+(n−3)d]−k×((n−1)/2)[2a+(n−2)d] d=((2a)/2)[n+n−2−k(n−1)]+(d/2)[n^2 −n+n^2 −5n+6−k(n−1)(n−2)] d=a[2n−2−kn+k]+(d/2)[2n^2 −6n+6−kn^2 +3kn−2k] d=a[2n−kn+k−2]+(d/2)[2n^2 −kn^2 +3kn−6n+6−2k] so 2n−kn+k−2=0 k(1−n)=2−2n k=((2(1−n))/(1−n))=2

Sn=n2[2a+(n−1)d]Sn−1=n−12[2a+(n−2)d]Sn−2=n−22[2a+(n−3)d]d=n2[2a+(n−1)d]+n−22[2a+(n−3)d]−k×n−12[2a+(n−2)d]d=2a2[n+n−2−k(n−1)]+d2[n2−n+n2−5n+6−k(n−1)(n−2)]d=a[2n−2−kn+k]+d2[2n2−6n+6−kn2+3kn−2k]d=a[2n−kn+k−2]+d2[2n2−kn2+3kn−6n+6−2k]so2n−kn+k−2=0k(1−n)=2−2nk=2(1−n)1−n=2

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