设{an}是等差数列,bn=(1/2)an.已知b1+b2+b3=21/8,b1b2b3=1/8.求等差数列的通项an.

问题描述:

设{an}是等差数列,bn=(

1
2
an.已知b1+b2+b3=
21
8
,b1b2b3=
1
8
.求等差数列的通项an

设等差数列{an}的公差为d,则an=a1+(n-1)d.∴bn=(12)a1+(n-1)db1b3=(12)a1•(12)a1+2d=(12)2(a1+d)=b22.由b1b2b3=18,得b23=18,解得b2=12.代入已知条件b1b2b3=18b1+b2+b3=218.整理得b1b3=14b1+b3=178.解这个方...