问题 解答题
已知{an}是各项均为正数的等差数列,lga1,lga2,lga4成等差数列.又bn=
1
a2n
,n=1,2,3,….
(Ⅰ)证明{bn}为等比数列;
(Ⅱ)如果数列{bn}前3项的和等于
7
24
,求数列{an}的首项a1和公差d.
答案

(1)证明:设{an}中首项为a1,公差为d.

∵lga1,lga2,lga4成等差数列

∴2lga2=lga1+lga4

∴a22=a1?a4,即(a1+d)2=a1(a1+3d)

∴d=0或d=a1

当d=0时,an=a1,bn=

1
a2n
=
1
a1

bn+1
bn
=1,

∴{bn}为等比数列;

当d=a1时,an=na1,bn=

1
a2n
=
1
2na1

bn+1
bn
=
1
2

∴{bn}为等比数列

综上可知{bn}为等比数列

(2)当d=0时,bn=

1
a2n
=
1
a1

∴b1+b2+b3=

3
a1
=
7
24

∴a1=

72
7

当d=a1时,bn=

1
a2n
=
1
2na1

∴b1+b2+b3=

1
2a1
+
1
4a1
+
1
8a1
=
7
8a1
=
7
24

∴a1=3

综上可知

a1=
72
7
d=0
a1=3
d=3

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