解题思路:(Ⅰ)由已知条件推导出2a1qm+1=a1qm+a1qm-1,由此能求出q的值.
(Ⅱ)若q=1,由a1≠0,得2Sm+2≠Sm+Sm+1;q=-[1/2],能推导出2Sm+2=Sm+Sm+1.故当q=1时,Sm,Sm+2,Sm+1不成等差数列;q=-[1/2]时,Sm,Sm+2,Sm+1成等差数列.
(Ⅰ)依题意,得2am+2=am+1+am,
∴2a1qm+1=a1qm+a1qm-1
在等比数列{an}中,a1≠0,q≠0,
∴2q2=q+1,解得q=1或-[1/2].
(Ⅱ)若q=1,Sm+Sm+1=ma1+(m+1)a1=(2m+1)a1,Sm+2=(m+2)a1
∵a1≠0,∴2Sm+2≠Sm+Sm+1
若q=-[1/2],Sm+2=
1−(−
1
2)m+2
1−(−
1
2)•a1=[[2/3−
1
6•(−
1
2)m]•a1,
Sm+Sm+1=
1−(−
1
2)m
1−(−
1
2)•a1+
1−(−
1
2)n+1
1−(−
1
2)•a1
={
4
3−
2
3]•[(-[1/2])m+(-[1/2])n+1}•a1,
∴2Sm+2=Sm+Sm+1,
∴2Sm+2=Sm+Sm+1,
∴当q=1时,Sm,Sm+2,Sm+1不成等差数列;
q=-[1/2]时,Sm,Sm+2,Sm+1成等差数列.
点评:
本题考点: 等比数列的性质;等差关系的确定;等差数列的性质.
考点点评: 本题考查等比数列的公比的求法,考查等差数列的判断,是中档题,解题时要认真审题,注意等差数列的性质的灵活运用.