Question

设等差数列{an}的前n项和为Sn，且a2= 4，S5=35，求{an}的前n项和Sn

Answer 1

a2=a1+d=4...............................................1
sn=na1+n(n-1)d/2,s5=5a1+10d=35...............2
1、2联立得a1=1,d=3
sn=na1+n(n-1)d/2=n+3n(n-1)/2=(3n平方-n)/2

Answer 2

①s5=1/2[5(a1+a5)]=35 -> 5a1+5a5=70
②a2=a1+(n-1)d=4 -> a1=4-d
将②带入① 得 20-5d+5a5=70 -> a5-d=10
又因为a5=a1+4d 将其带入a5-d=10 得 ③ a1+3d=10
将②带入③ 解得a1=1 d=3
sn=n/2(3n-1)

Answer 3

a2=a1+d=4
s5=5a1+10d=35
所以，a1=1,d=3
所以,an=1+3*(n-1)
sn=n+3*(n-1)*n*1/2

Answer 4

解：依题意，a1+d=4,5a1+5*（5-1）/2=35,得到a1+d=4,a1+2d=7，所以a1=1、d=3，故Sn=n*1+3n*(n-1)/2=n(3n-1)/2

Answer 5

a1+d=4 5a1+10d=35
则a1=1 d=3 an=3n-2
Sn=(3n-1)*n/2