Question

求等比数列的通项公式：设﹛An﹜是各项为正数且递增的等比数列,a1+a2+a3=14／3,1／a1+1／a2+1／a3=

Answer 1

a1+a2+a3=a1(1+q+q^2)=14/3 (1)

1/a1+1/a2+1/a3=1/a1*(1+1/q+1/q^2)=1/a1*(q^2+q+1)/q^2=21/8 (2)

(1)/(2)得(a1q)^2=16/9

则a2=4/3

a1+a2+a3=a2(1/q+1+q)=14/3

则q=1/2或q=2

由于数列是递增数列,则有q>1

所以有q=2.

a1=2/3

an=2/3*2^(n-1)