Question

等差数列{an}的前n项和为18,其中前三项和为1,末三项和为3,求项数n

Answer 1

设首项为a1 公差为d

3a1+3d=1 3a1+(3n-6)d=3 可得2a1=2/3-2d

相减得（3n-9）d=2

[a1+a1+(n-1)d]*n/2=18

[2a1+(n-1)d]*n=36

[2/3-2d+(n-1)d]*n=36 [2+(3n-9)d]*n=108

4n=108 n=27