已知SN是等差数列,S6=36,SN=324,S(N-6)=144,则N=?

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已知SN是等差数列,S6=36,SN=324,S(N-6)=144,则N=?
已知SN是等差数列,S6=36,SN=324,S(N-6)=144,则N=?

已知SN是等差数列,S6=36,SN=324,S(N-6)=144,则N=?
Sn-S
=a+a+……+an
=324-144
=180
S6=a6+a5+……+a1=36
两式相加得
(a+a6)+(a+a5)+……+(an+a1)=180+36
6(a1+an)=216
a1+an=36
Sn=n(a1+an)/2=324
n*36/2=324
n=18

设首相为a,公差为d,
由题意可得:
2a+5d=12--------(1)
2na+(n-1)nd=648-------(2)
2(n-6)a+(n-7)(n-6)d=288------(3)
然后:(3)+(1)*2=2na+{(n-7)(n-6)+30}d=360-------(4)
(2)-(4)=(12n-72)d=288
化简:(...

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设首相为a,公差为d,
由题意可得:
2a+5d=12--------(1)
2na+(n-1)nd=648-------(2)
2(n-6)a+(n-7)(n-6)d=288------(3)
然后:(3)+(1)*2=2na+{(n-7)(n-6)+30}d=360-------(4)
(2)-(4)=(12n-72)d=288
化简:(n-6)d=24
(3)-(1)*(n-6)=(n-12)(n-6)d=288-12n+72
化简得:24n-288=288-12n+72
即n=18
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等差数列前n项和

Sn - S(n-6) = 324 - 144
a(n-5) + a(n-4) + a(n-3) + a(n-2) + a(n-1) + an = 180
a1 + a2 + a3 + a4 + a5 + a6 = 36

两式子相减
(a(n-5) - a1) + (a(n-4) - a2) + …… + (an - a6) = 180 - 36
(n-6)d + (n-6)d + (n-6)d + (n-6)d + (n-6)d + (n-6)d = 144
6(n-6)d = 144
(n-6)d = 24

S6 = 36
6a1+15d = 36
2a1 = 12 - 5d

Sn = 324
[2a1 + (n-1)d] n /2 =324

[12 + (n-6)d]  n  = 648
(12 + 24)*n = 648
n = 648/36 = 18


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