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](https://www.tinkutara.com/question/Q55861.png)
$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$${S}_{{n}−\mathrm{1}} =\frac{{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{2}\right){d}\right] \\ $$$${S}_{{n}−\mathrm{2}} =\frac{{n}−\mathrm{2}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{3}\right){d}\right] \\ $$$${d}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]+\frac{{n}−\mathrm{2}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{3}\right){d}\right]−{k}×\frac{{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{2}\right){d}\right] \\ $$$${d}=\frac{\mathrm{2}{a}}{\mathrm{2}}\left[{n}+{n}−\mathrm{2}−{k}\left({n}−\mathrm{1}\right)\right]+\frac{{d}}{\mathrm{2}}\left[{n}^{\mathrm{2}} −{n}+{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{6}−{k}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\right] \\ $$$${d}={a}\left[\mathrm{2}{n}−\mathrm{2}−{kn}+{k}\right]+\frac{{d}}{\mathrm{2}}\left[\mathrm{2}{n}^{\mathrm{2}} −\mathrm{6}{n}+\mathrm{6}−{kn}^{\mathrm{2}} +\mathrm{3}{kn}−\mathrm{2}{k}\right] \\ $$$${d}={a}\left[\mathrm{2}{n}−{kn}+{k}−\mathrm{2}\right]+\frac{{d}}{\mathrm{2}}\left[\mathrm{2}{n}^{\mathrm{2}} −{kn}^{\mathrm{2}} +\mathrm{3}{kn}−\mathrm{6}{n}+\mathrm{6}−\mathrm{2}{k}\right] \\ $$$${so}\:\:\mathrm{2}{n}−{kn}+{k}−\mathrm{2}=\mathrm{0} \\ $$$${k}\left(\mathrm{1}−{n}\right)=\mathrm{2}−\mathrm{2}{n} \\ $$$${k}=\frac{\mathrm{2}\left(\mathrm{1}−{n}\right)}{\mathrm{1}−{n}}=\mathrm{2} \\ $$