Question

已知an是一个公差大于零的等差数列且满足a三a五等于45,a2+a6=14.求an的通��？

Answer 1

an=a1+(n-1)d
a2+a6=14
2a1+6d=14
a1+3d=7 (1)
a3.a5=45
(a1+2d)(a1+4d)=45
(a1+3d-d)(a1+3d+d) =45
(7-d)(7+d)=15
49-d^2=15
d^2=34
d=√34 or -√34(rej)
from (1)
a1+3d=7
a1+3√34=7
a1=7-3√34
an
= a1+(n-1)d
=7-3√34 +√34.(n-1)
=√34.n +7 -4√34

Answer 2

由题意可以，a3+a5=a2+a6=14，a3*a5=45，设公差为d>0，则有2a3+2d=14，a3+d=7，a3=7-d，a3(a3+2d)=45，代入后式得，(7-d)(7+d)=49-d²=45，解得d=2(舍去负值)。则a3=5，a1=1，an=2n-1。