Question

在等比数列{an}中,若a4-a2=24,a2+a3=6,求首项a1和公比q.这道题怎么做?

Answer 1

{an}为等比数列，
a4-a2=24即a2(q^2- 1) =24;即a2(q+1)(q-1)=24.......①
a2+a3=6即a2(1+q)=6.........②
将②代入①可得6(q-1)=24, q=5;
将q=5代入②，得a2=a1q=1;得a1=1/5

Answer 2

a4 = a2 * q*q a3=a2*q
所以a4-a2=24 a2+a3=6
即a2(q*q-1) = 24
a2(1+q) = 6
解得a2 =1 q= 5
所以a1 = 1/5 q=5

Answer 3

a4-a2=a1q^3-a1q=a1q(q^2-1)=24
a2+a3=a1q+a1q^2=a1q(q+1)=6
相除
(q^2-1)/(q+1)=4
q-1=4
q=5

代入a1q(q+1)=6
a1=1

Answer 4

a1=1/5 q=5因为 （a4-a2)/(a2+a3)=q-1=4所以 q=5 a1=1/5