Question

在等差数列中,若Sn=156,A(n-5)=30,S11=99，求n？

Answer 1

设公差为D,初始值为A(1)
则S11=A(11)+A(10)+....+A(1)=A(1)+10D+A(1)+9D+....+A(1)+D+A(1)
=11A(1)+(1+10)*10D/2=11(A(1)+5D)
==>A(1)+5D=99/11=9 (1) ==>A(1)=9-5D
A(N-5)=A(1)+(N-6)D=30 (2)
(2)-(1): (N-11)D=21 ==>D=21/(N-11)

则SN=NA(1)+N*(N-1)D/2=156
N(A(1)+(N-1)D/2)=156
N(9-5D+(N-1)D/2)=156
N(18-10D+ND-D)=312
N(18+(N-11)*21/(N-11))=312
N(18+21)=312
N=312/39=8

Answer 2

S11=99
==》 (a1+a11)*11/2=99
a1+a11=18
==> a6=9
Sn=(a1+an)*n/2
=(a(n-5)+a6)*n/2
=(30+9)*n/2
=156
==> n=8

Answer 3

n=8