已知公差大于零的等差数列{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b
(1)求数列{an}的通项公式an;
(2)若数列{bn}是等差数列,且b
(1)an为等差数列,a3•a4=117,a2+a5=22
又a2+a5=a3+a4=22
∴a3,a4是方程x2-22x+117=0的两个根,d>0
∴a3=9,a4=13
∴
a1+2d=9
a1+3d=13
∴d=4,a1=1
∴an=1+(n-1)×4=4n-3
(2)由(1)知,sn=n+
n(n−1)×4
2=2n2−n
∵bn=
sn
n+c=
2n2−n
c+n
∴b1=
1
1+c,b2=
6
2+c,b3=
15
3+c,
∵bn是等差数列,∴2b2=b1+b3,∴2c2+c=0,
∴c=−
1
2(c=0舍去)
又a2+a5=a3+a4=22
∴a3,a4是方程x2-22x+117=0的两个根,d>0
∴a3=9,a4=13
∴
a1+2d=9
a1+3d=13
∴d=4,a1=1
∴an=1+(n-1)×4=4n-3
(2)由(1)知,sn=n+
n(n−1)×4
2=2n2−n
∵bn=
sn
n+c=
2n2−n
c+n
∴b1=
1
1+c,b2=
6
2+c,b3=
15
3+c,
∵bn是等差数列,∴2b2=b1+b3,∴2c2+c=0,
∴c=−
1
2(c=0舍去)
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