Question

设Sn是公差不为0的等差数列{an}的前n项和，若a1=2a8-3a4，则S8/S16=

Answer 1

前n项和公式：Sn=na1+n(n-1)d/2
通项公式：an=a1+(n-1)d
故a8=a1+7d,a4=a1+3d,代入已知条件得：a1=5d/2
S8=8a1+28d (1)
S16=16a1+120d (2)
a1=5d/2 (3)
最后结果你算下就好了，二元二次方程组，今天翻了下机械设计手册看见了，就解答下，希望你还满意。

Answer 2

an = a1+(n-1)d
Sn = (2a1+(n-1)d)n/2

a1= 2a8-3a4
= -a1 +5d
a1= 5d/2 (1)

S8/S16 = (2a1+7d)/[2(2a1+15d)]
= (12d)/(40d)
= 3/10

Answer 3

a1=2a8-3a4
a1=2a1+14d-3a1-9d
2a1=5d
S8/S16=(a1+a8)*8/(a1+a16)*16
=(2a1+7d)/2(2a1+15d)
=12d/40d
=3/10