问题 解答题
设首项为a,公差为d的等差数列前n项的和为An,又首项为a,公比为r的等比数列前n项和为Gn,其中a≠0,|r|<1.令Sn=G1+G2+…+Gn,若有
lim
n→∞
(
An
n
-Sn)
=a,求r的值.
答案

由题意知Gn=

a(1-rn)
1-r

∴Sn=

1
-1+r
•[a(r +r2+r3…+rn)-(a+a+a…+a)]

=

1
-1+r
(
ar(1-rn)
1-r
-na)

=

a
(-1+r)2
[rn-r-n(-1+r)]

An=na+

n(n-1)
2
•d

An
n
-Sn=
1
n
[na+
n(n-1)
2
•d
]-
a
(-1+r)2
[rn-r-n(-1+r)]=a+
n-1
2
•d
-
a
(-1+r)2
×(rn-r)-
an
1-r

lim
n→∞
(
An
n
-Sn)=a,a≠0,|r|<1

所以:

d
2
+
a
r-1
=0且
a
(1-r)2
×r+a-
d
2
=a,即
a
(1-r)2
×r-
d
2
=0

a
(1-r)2
×r+
a
r-1
=0,整理得2r-1=0,解得r=
1
2

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