Question

设Sn是等差数列{an}的前n项和,已知S6=36,Sn=324,Sn-6=144

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

Answer 1

Sn - S<n-6> = 324 - 144
a<n-5> + a<n-4> + a<n-3> + a<n-2> + a<n-1> + a<n> = 180

a1 + a2 + a3 + a4 + a5 + a6 = 36

两式子相减
(a<n-5> - a1) + (a<n-4> - a2) + …… + (a<n> - 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
(2a1 + 5d)*6/2 = 36
2a1 = 12 - 5d

Sn = 324
[2a1 + (n-1)d] *n /2 =324
(12 - 6d + nd) * n = 648
[12 + (n-6)d] * n = 648
(12 + 24)*n = 648
n = 648/36 = 18

Answer 2

n=18,an=1,3,5,7,...,35.

Answer 3

n = 648/36 = 18