在等差数列{an}中,前3项和为9,后3项和为54,各项和为105,则n=

an=a1+(n-1)d
前3项和为9
a1+a2+a3=9
3a1+3d=9
a1+d =3 (1)
后3项和为54
a(n-2)+a(n-1)+an =54
3a1+(3n-6)d =54
a1+(n-2)d = 18
3-d + (n-2)d = 18 ( from (1) :a1=3-d )
(n-3)d =15
d = 15/(n-3) (2)
各项和为105
a1+a2+...+an =105
(2a1+(n-1)d)n =210
( 2[3- 15/(n-3)] + (n-1)[15/(n-3)] )n =210
[6(n-3)-30+15(n-1)]n = 210(n-3)
(21n-63)n = 210(n-3)
(n-3)n = 10(n-3)
n^2-13n+30=0
(n-10)(n-3)=0
n=10 or 3( rejected)
ie n=10