已知等差数列an满足an>0,a5+a9+a11=15,且a5 +2,a9 +5,a13 +13分别是等比数列

题目

已知等差数列an满足an>0,a5+a9+a11=15,且a5 +2,a9 +5,a13 +13分别是等比数列
等比数列bn中的第三项,第四项和第五项,求bn通项
数列bn的前n项和为Sn，求证：数列 Sn +4/5 是等比数列

答案

a5+a9+a11=15
3a9=15
a9=5
(a9+5)^2=(a5+2)(a13+13)
(a9+5)^2=a5*a13+13a5+2a13+26
(a9)^2+10a9+25=(a9-4d)(a9+4d)+13(a9-4d)+2(a9+4d)+26
(a9)^2+10a9+25=(a9)^2-16d^2+13a9-52d+2a9+8d+26
10a9+25=-16d^2+15a9-44d+26
-16d^2+5a9-44d+1=0
-16d^2+5*5-44d+1=0
-16d^2-44d+26=0
8d^2+22d-13=0
(2d-1)(4d+13)=0
d=1/2或d=-13/4(舍去)
b3=a5+2
=a9-4d+2
=5-4*1/2+2
=5
b4=a9+5
=5+5
=10
b5=a13+13
=a9+4d+13
=5+4*1/2+13
=20
q=b4/b3=10/5=2
bn=b1q^(n-1)
=b3q^(n-3)
=5*2^(n-3)