求等差数列``

a4=-3 a9=12

d=(a9-a4)/(9-4)=3

a1=a4-3d=-12

S12=12(a1+a12)/2=6(a4+a9)=54

等差数列an的等差是
（a9-a4）/5=3
所以a1= a4-3*3=-3-9=-12
a12=a9+3*3=12+9=21
S12=（a1+a12）*6=54

a1+3d=-3
a1+8d=12z这两个式子连列。求出a1=-12和d=3
s12=（a1+a12）/2=(-12-12+33)/2=4.5

d=[12-(-3)]/(9-4)=3
a1=-3-2d=-9
a1+a12=a4+a9=9
S12=(a1+a12)12/2=54