1.
设{an}公差为d,则d≠0,设{bn}公比为q.a5,a8,a13是等比数列连续三项,则
a8²=a5×a13
(a1+7d)²=(a1+4d)(a1+12d)
整理,得
d²-2a1d=0
d(d-2a1)=0
d=0(舍去)或d=2a1
q=a8/a5=(a1+7d)/(a1+4d)=15d/(9d)=5/3
b1=b2/q=5/(5/3)=3
bn=b1q^(n-1)=3×(5/3)^(n-1)=5^(n-1)/3^(n-2)
数列{bn}的通项公式为bn=5^(n-1)/3^(n-2)
2.
设{an}的公差为d,则d≠0
a1,a3,a9成等比,则
a3²=a1×a9
(a1+2d)²=a1(a1+8d)
整理,得
d²-a1d=0
d(d-a1)=0
d=0(舍去)或d=a1
an=a1+(n-1)d=a1+(n-1)a1=na1
(a1+a3+a9)/(a2+a4+a10)=(a1+3a1+9a1)/(2a1+4a1+10a1)=13a1/(16a1)=13/16