公差不为零的等差数列{an}中,已知其前n项和为Sn,若S8=S5+45,且a4,a7,a12成等比数列(Ⅰ)求数列{an}的通项an(Ⅱ)当bn=1Sn时,求数列{bn}的前n和Tn.-数学

题目简介

公差不为零的等差数列{an}中,已知其前n项和为Sn,若S8=S5+45,且a4,a7,a12成等比数列(Ⅰ)求数列{an}的通项an(Ⅱ)当bn=1Sn时,求数列{bn}的前n和Tn.-数学

题目详情

公差不为零的等差数列{an}中,已知其前n项和为Sn,若S8=S5+45,且a4,a7,a12成等比数列
(Ⅰ)求数列{an}的通项an
(Ⅱ)当bn=
1
Sn
时,求数列{bn}的前n和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由S8=S5+45得,S8-S5=45,
∴a6+a7+a8=45,即3a7=45,得a7=15,
又∵a72=a4a12,设公差为d≠0,
a1+6d=15
(a1+6d)2=(a1+3d)(a1+11d)

解得
a1=3
d=2

∴an=2n+1,
(Ⅱ)由(Ⅰ)得Sn=
n(3+2n+1)
2
=n(n+2)

bn=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

Tn=b1+b2+b3+…+bn
=class="stub"1
2
[(class="stub"1
1
-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+2
)]
=class="stub"1
2
(class="stub"1
1
+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2
)

Tn=class="stub"3
4
-class="stub"2n+3
2(n+1)(n+2)

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