Question

等差数列-1,2,5,8,...前多少项和为125

Answer 1

设公差为d

d=2-(-1)=5-2=8-5=...=3

Sn=na1+n(n-1)d/2=-n+3n(n-1)/2=(3n²-5n)/2

令Sn=125

(3n²-5n)/2=125

整理,得3n²-5n-250=0

(3n+25)(n-10)=0

n=-25/3(n为正整数,舍去)或n=10

此等差数列的前10项和为125.