已知等差数列{an}中,a3+a6=17,a1a8=-38,且a1

(1)
an = a1+(n-1)d
a3+a6=17
2a1+7d=17 (1)
a1a8=-38
a1(a1+7d) = -38 (2)
sub (1) into (1)
a1(a1+(17-2a1) ) =-38
a1^2-17a1-38=0
a1= (17+21)/2 or (17-21)/2
a1=39/2(rejected) or -2
a1=-2 d= 3
an = -2+(n-1)3 = -5+3n
(2)
a1=-2
a2= -2+3= 1
a3= -2+6 = 4
b1=a2
b2=a1
b3=a3
q= -2
b1+b2+b3 = -2+1+4= 3