1在等差数列{an}中,a8-1/2a11=6则数列前9项之和=?A24B48C72D1082.在等比数列{an}中.a1=2.前n项和为Sn,若数列{an+1}也是等比数列.则Sn=?

Question

1在等差数列{an}中,a8-1/2a11=6则数列前9项之和=?
A 24 B48 C 72 D108
2.在等比数列{an}中.a1=2.前n项和为Sn,若数列{an+1}也是等比数列.则Sn=?

Answer 1

1.用通项公式an=a1+(n-1)d将条件展开得a5=12,所以S9=9a5=108,选D
2.a[n]=2q^(n-1)
{a[n]+1}是等比数列,则(a[n-1]+1)(a[n+1]+1)=(a[n]+1)^2
即（2q^(n-2)+1)*(2q^n+1)=(2q^(n-1)+1)^2
解得q=1
故an=2,Sn=2n