已知公差大于零的等差数列{a
n}的前n项和为S
n,且满足:a
3•a
4=117,a
2+a
5=22.
(1)求数列{a
n}的通项公式a
n;
(2)若数列{b
n}是等差数列,且
bn=,求非零常数c.
人气:456 ℃ 时间:2019-11-04 09:09:10
解答
(1)a
n为等差数列,a
3•a
4=117,a
2+a
5=22
又a
2+a
5=a
3+a
4=22
∴a
3,a
4是方程x
2-22x+117=0的两个根,d>0
∴a
3=9,a
4=13
∴
∴d=4,a
1=1
∴a
n=1+(n-1)×4=4n-3
(2)由(1)知,
sn=n+=2n2−n∵
bn==∴
b1=,
b2=,
b3=,
∵b
n是等差数列,∴2b
2=b
1+b
3,∴2c
2+c=0,
∴
c=−(c=0舍去)
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