Question

设Sn是等差数列{an}的前n项和，已知S6=36，Sn=324,S(n-6)=144, (n〉6)则n等于？

Answer 1

Sn是等差数列
S6=a1*6+6(6-1)/2*d=36，则2a1+5d=12......&
最后六项的和S=an*6-6(6-1)/2*d=6an-15d
S(n-6)=Sn-S=324-(6an-15d)=144,则2an-5d=60......@
&+@:a1+an=36
Sn=(a1+an)/2*n
n=18

Answer 2

Sn－S(n-6)＝180为后6项的和,S6为前6项的和，
把他们相加再除以6得到a1+an＝36（等差数列性质）
所以Sn=n×（a1+an）/2＝324
n＝18

Answer 3

此题不对

Answer 4

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