d=3
an=a1+(n-1)d=11
即a1+3n-3=11 a1+3n=14 a1=14-3n (1)
Sn=[2a1+(n-1)d]*n/2=14
即(2a1+3n-3)*n=28 (2)
(1)代入(2) (28-6n+3n-3)*n=28
即-3n²+25n-28=0
3n²-25n+28=0
(3n-4)(n-7)=0
解得n=4/3(舍去)或n=7
代入(1)a1=14-3*7=-7
所以a3=a1+2d=-7+2*3=-1