在公差不为0的等差数列{an}中,a1,a4,a8成等比数列.(1)已知数列{an}的前10项和为45,求数列{an}的通项公式;(2)若bn=1anan+1,且数列{bn}的前n项和为Tn,若Tn=

题目简介

在公差不为0的等差数列{an}中,a1,a4,a8成等比数列.(1)已知数列{an}的前10项和为45,求数列{an}的通项公式;(2)若bn=1anan+1,且数列{bn}的前n项和为Tn,若Tn=

题目详情

在公差不为0的等差数列{an}中,a1,a4,a8成等比数列.
(1)已知数列{an}的前10项和为45,求数列{an}的通项公式;
(2)若bn=
1
anan+1
,且数列{bn}的前n项和为Tn,若Tn=
1
9
-
1
n+9
,求数列{an}的公差.
题型:解答题难度:中档来源:不详

答案

设等差数列{an}的公差为d,由a1,a4,a8成等比数列可得,a42=a1a8
(a1+3d)2=a1(a1+7d)
a12+6a1d+9d2=a12+7a1d,而d≠0,∴a1=9d.
(1)由数列{an}的前10项和为45,得S10=10a1+class="stub"10×9
2
d=45

即90d+45d=45,故d=class="stub"1
3
,a1=3,
故数列{an}的通项公式为an=3+(n-1)•class="stub"1
3
=class="stub"1
3
(n+8)

(2)bn=class="stub"1
anan+1
=class="stub"1
d
(class="stub"1
an
-class="stub"1
an+1
)

则数列{bn}的前n项和为Tn=class="stub"1
d
[(class="stub"1
a1
-class="stub"1
a2
)+(class="stub"1
a2
-class="stub"1
a3
)+…+(class="stub"1
an
-class="stub"1
an+1
)]

=class="stub"1
d
(class="stub"1
a1
-class="stub"1
an+1
)=class="stub"1
d
(class="stub"1
9d
-class="stub"1
9d+nd
)=class="stub"1
d2
(class="stub"1
9
-class="stub"1
n+9
)
=class="stub"1
9
-class="stub"1
n+9

故数列{an}的公差d=1或d=-1.

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