Question

数列{an}是公差不为零的等差数列,且a7,a10,a15是等比数列{bn}的连续三项,若该等比数列的首项b1＝3,则bn...

数列{an}是公差不为零的等差数列,且a7,a10,a15是等比数列{bn}的连续三项,若该等比数列的首项b1＝3,则bn等于?

Answer 1

解：
设等差数列{an}的公差为d，则由题意可知：
(a10)^2＝a7×a15
即(a7＋3d)^2＝a7×(a7＋8d)
解得a7＝9/2d
∴数列{bn}的公比q＝a10/a7＝(a7＋3d)/a7＝5/3
又b1＝3
∴bn＝b1*q^(n－1)＝3*(5/3)^(n－1)

Answer 2

a10²=a7a15
(a1+9d)²=(a1+6d)(a1+14d)
a1²+18a1d+81d²=a1²+20a1d+84d²
2a1d=-3d²
所以a1=-3d/2

q=a10/a7=(a1+9d)/(a1+6d)=(15d/2)/(9d/2)=5/2
b1=3
所以bn=3*(5/2)^(n-1)

Answer 3

设a7=a1+6d,a10=a1+9d,a15=a1+14d
a7,a10,a15是等比数列{bn}的连续三项(a10)^2=a7*a15
(a1+9d)^2=(a1+6d)*(a1+14d)整理得a1=(-3/2)d,a10=(15/2)d,a7=(9/2)d
a=a10/a7=5/3
所以bn=b1*q^(n-1)=3*(5/3)^(n-1)