设a1,d为实数,首项为a1,公差为d的等差数列{an}的前n项和为Sn,满足S5+S6+15=0.

S5 + S6 + 15 = 0
5 + S6 + 15 = 0
S6 = -20
S5 = a1+a2+a3+a4+a5
= a1+(a1+d)+(a1+2d)+(a1+3d)+(a1+4d)
= 5*a1 + 10*d
= 5
a1 + 2d = 1
d = (1-a1)/2
S6 = a1+a2+a3+a4+a5+a6
= a1+(a1+d)+(a1+2d)+(a1+3d)+(a1+4d)+(a1+5d)
= 6*a1 + 15*d
= -20
6*a1 + 15d = -20
6*a1 + 15(1-a1)/2 = -20
12*a1 + 15 - 15*a1 = -40
55 = 3*a1
a1 = 55/3
d = (1 - 55/3)/2 = -26/3

（1）S5+S6+15=0.，S5=5
S6=-20 a6=S6-S5=-25
（a1+a6）*6/2=-20 a1+a6=-20/3 a1=-55/3
第二题我再算下