已知等差数列{an}中,a2=-4,a5=a3+4,等比数列{bn}以a5-1,a6-1,a8+1为前三项.求：1）数列

a5=a3+4
a2+3d=a2+d+4
2d=4
d=2
a2=a1+d
-4=a1+2
a1=-6
an=a1+(n-1)d
=-6+2(n-1)
=2n-8
a10=2*10-8=12
Sn=(a1+an)*n/2
=[-6+2n-8]*n/2
=n(n-7)
b1=a5-1
=2*5-8-1
=1
b2=a6-1
=2*6-8-1
=3
b3=a8+1
=2*8-8+1
=9
q=b2/b1=3/1=3
bn=b1q^(n-1)
=3^(n-1)
Sbn=b1(1-q^n)/(1-q)
=(1-3^n)/(1-3)
=(3^n-1)/2

（1）an=a1+（n-1）d
a5=a3+2d，所以d=2，a1=-6
a10=a1+9d=12
an=-6+2（n-1）=2n-8
（2）a5-1=b1=1 a6-1=b1q=3,q=3
bn=b1q^(n-1)=1*3^(n-1)=1/3*3^n

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