已知{an}是公比为q的等比数列,且am、am+2、am+1成等差数列.

2个回答

  • 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.