问题描述: 设Sn为等差数列{an}的前n项和,已知s6=36,Sn=324 ,S(n-6)=144 ,(n>6) ,求n的值 1个回答 分类:数学 2014-11-21 问题解答: 我来补答 等差数列前n项和Sn=na1 +n*(n-1)*d/2n=6时S6=6a1 +6*5*d/2S6=6a1 +15d36=6a1 +15da1=6-(5/2)dSn=na1 +n*(n-1)*d/2=324将a1代入6n-5nd/2 +n*(n-1)*d/2=3246n + n[n-6]d/2=324d/2 = (324-6n)/[n(n-6)],S(n-6)=[(n-6)a1 +(n-6)*(n-7)*d/2]=144(6-5d/2)(n-6)+n(n-6)d/2 - 7(n-6)d/2=144将上面求得的d/2代入化简得到n=18 展开全文阅读