等差数列的公差为3,若a2,a4,a8成等比数列,求a4

问题描述:

等差数列的公差为3,若a2,a4,a8成等比数列,求a4

公差为3
a2=a4-6
a8=a4+12
a2,a4,a8成等比数列
所以
a2*a8=a4^2
所以a4^2=(a4-6)(a4+12)
解得,a4=12

由题意:a2=a4-6 , a8=a4+12
a2,a4,a8成等比数列 ,则:(a4)^2=a2*a8
(a4)^2 = (a4-6)*(a4+12)
(a4)^2 = (a4)^2 + 6a4 - 72
a4=12

设a2=a4-6
a8=a4+12
由于a2,a4 ,a8成等比数列
a4*a4=a2*a8解得:a4*a4=(a4-6)(a4+12)
解得:72=2a4
a4=12

a2=a4-6, a8=a4+12. 所以a4/a2=a8/a4,把上面的a2和a8带入,计算得出a4=12

a2=a1+3,
a4=a1+9,
a8=a1+21
a4/a2=a8/a4
(a1+9)/(a1+3)=(a1+21)/(a1+9)
a1=3
a4=3+9=12