已知公差数列大于零的等差数列{an}的前n项和为Sn,且满足a3*a4=117,a2+a5=22
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已知公差数列大于零的等差数列{an}的前n项和为Sn,且满足a3*a4=117,a2+a5=22
(1)求数列{an}的通项公式an;(2)若数列{bn}是等差数列,且bn=Sn/(n+c),求非零常数c
(1)求数列{an}的通项公式an;(2)若数列{bn}是等差数列,且bn=Sn/(n+c),求非零常数c
(1)∵d>0,且a2+a5=22
∴a3+a4=22
又∵a3*a4=117
解得a3=9,a4=13
d=a4-a3=4
∵a3=a1+2d
∴a1=1
∴an=a1+(n-1)d=1+4(n-1)=4n-3 Sn=a1+[n(n-1)*d]/2=n(2n-1)
(2)若{bn}为等差数列,且bn=Sn/(n + C)
∵{bn}为等差数列,则 b(n) =n(2n-1)/(n + C),
故 b(1) =1*(2*1-1)/(1 + C) =1/(1 + C) ..①
b(2) =2*(2*2-1)/(2 + C) =6/(2 + C).②
b(3) =3*(2*3-1)/(3 + C) =15/(3 + C) .③
根据等差数列性质:
b(2)-b(1) =b(3) -b(2) 即2b(2)=b(1) +b(3)
①、②、③代入上式得
12/(2 + C) =1/(1 + C) =15/(3 + C)
整理后得:4C^2+2C=0
解得 C=0 (舍去)
C=-1/2 (唯一解)
期待您的采纳
∴a3+a4=22
又∵a3*a4=117
解得a3=9,a4=13
d=a4-a3=4
∵a3=a1+2d
∴a1=1
∴an=a1+(n-1)d=1+4(n-1)=4n-3 Sn=a1+[n(n-1)*d]/2=n(2n-1)
(2)若{bn}为等差数列,且bn=Sn/(n + C)
∵{bn}为等差数列,则 b(n) =n(2n-1)/(n + C),
故 b(1) =1*(2*1-1)/(1 + C) =1/(1 + C) ..①
b(2) =2*(2*2-1)/(2 + C) =6/(2 + C).②
b(3) =3*(2*3-1)/(3 + C) =15/(3 + C) .③
根据等差数列性质:
b(2)-b(1) =b(3) -b(2) 即2b(2)=b(1) +b(3)
①、②、③代入上式得
12/(2 + C) =1/(1 + C) =15/(3 + C)
整理后得:4C^2+2C=0
解得 C=0 (舍去)
C=-1/2 (唯一解)
期待您的采纳
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