(文科)数列{an}是首项为21,公差d≠0的等差数列,记前n项和为Sn,若110S10和119S19的等比中项为116S16.数列{bn}满足:bn=anan+1an+2.求:(1)数列{an}的通

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(文科)数列{an}是首项为21,公差d≠0的等差数列,记前n项和为Sn,若110S10和119S19的等比中项为116S16.数列{bn}满足:bn=anan+1an+2.求:(1)数列{an}的通

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(文科)数列{an}是首项为21,公差d≠0的等差数列,记前n项和为Sn,若
1
10
S10
1
19
S19的等比中项为
1
16
S16.数列{bn}满足:bn=anan+1an+2
求:(1)数列{an}的通项an;(2)数列{bn}前n项和Tn最大时n的值.
题型:解答题难度:中档来源:不详

答案

(1)设an=21+(n-1)d(d≠0),
则Sn=21n+
n(n-1)
2
d,
class="stub"1
n
Sn=21+class="stub"n-1
2
d,
class="stub"1
10
S10=21+class="stub"9
2
d,class="stub"1
19
S19=21+9d,class="stub"1
16
S16=21+class="stub"15
2
d.
由题设可知:(class="stub"1
16
S16)2=(class="stub"1
10
S10)•(class="stub"1
19
S19),
即(21+class="stub"15
2
d)2=(21+class="stub"9
2
d)•( 21+9d),解得d=-2,
∴an=21-2(n-1)=23-2n;
(2)由an=23-2n>0,得n<12.
∴当n<10时,bn=anan+1an+2>0;
当n>11时,bn=anan+1an+2<0.
而Tn=Tn-1+bn,
∴当bn>0时,Tn>Tn-1;当bn<0时,Tn<Tn-1.
∴当n<10时,{Tn}递增;当n>11时,{Tn}递减.
又b10=a10a11a12=-3,b11=a11a12a13=3,
∴T9=T11,
∴当n=9或11时,{ Tn}取最大值.

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