题目内容

若a、b、c成等比数列,logca、logbc、logab成等差数列,则公差d=
 
考点:等比数列的性质
专题:等差数列与等比数列
分析:设公比为k,可得logca=
1
1+2
lgk
lga
,logbc=
2
lgk
lga
+1
lgk
lga
+1
,logab=
lgk
lga
+1,由成等差数列可得x的方程,解得x=0或x=
3+
33
4
或x=
3-
33
4
,分别代入求公差即可.
解答: 解:∵a、b、c成等比数列,可设其公比为k,
∴b=ak,c=ak2
∴logca=
lga
lgc
=
lga
lgak2
=
lga
lga+2lgk
=
1
1+2
lgk
lga

同理可得logbc=
2
lgk
lga
+1
lgk
lga
+1
,logab=
lgk
lga
+1,
又∵logca、logbc、logab成等差数列,设
lgk
lga
=x,
∴2
2x+1
x+1
=
1
1+2x
+x+1,整理可得2x3-3x2-3x=0,
解得x=0或x=
3+
33
4
或x=
3-
33
4

当x=0时,logca=1,logbc=1,公差d=logbc-logca=0;
当x=
3+
33
4
时,logca=-
33
+5
4
,logbc=
1-
33
4
,d=logbc-logca=
3
2

当x=
3-
33
4
时,logca=
33
-5
4
,logbc=
33
+1
4
,d=logbc-logca=
3
2

故答案为:0或
3
2
点评:本题考查等比数列的性质和等差数列,涉及分类讨论的思想,属中档题.
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