a1+a2~a12

設x=a3, 等差數列公差=d,
(1) a1+a2+...+a5= 5*a3=5x=25, so, a3=x=5
(2) a1*a5=(x-2d)(x+2d)=-11, 25-4d^2=-11, so d=3 or -3
case 1 : d=3
a1=a3-2d=5-6= -1, a12=a1+11d= -1+33= 32
so, a1+..+a12= 6(a1+a12)= 6*31= 186
case 2: d=-3
a1=a3-2d=5+6=11, a12=a1+11d= 11- 33= -22
so, a1+...+a12= 6*(a1+a12)= 6*(-11)= -66

Ans: a1+a2+...+a12= 186 or -66

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設前五數為a-2d,a-d ,a,a+d,a+2d
==>a-2d+a-d+a+a+d+a+2d=25 ==>a=5

(a-2d)(a+2d)=-11 ==>d= +-3
所以前五數為-1,2,5,8,11 或11,8,5,2,-1
-1,2,5,----,32 ==>S12=(-1+32)*12/2=186

11,8,5,----,-22 ==>T12=(11+(-22))*12/2= -66


所以a1+a2~＋a12=186 或　-66