Question

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

Answer 1

等差数列前n项和Sn=na1 +n*（n-1）*d/2
n=6时
S6=6a1 +6*5*d/2
S6=6a1 +15d
36=6a1 +15d
a1=6-(5/2)d
Sn=na1 +n*(n-1)*d/2=324
将a1代入
6n-5nd/2 +n*(n-1)*d/2=324
6n + n[n-6]d/2=324
d/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