a2+a5=18,a3+a6=9,求an再问一个，已知等比数列an中，a1＝32公比q＝-1/2则a6＝？

Question

a2+a5=18,a3+a6=9,求an
再问一个，已知等比数列an中，a1＝32公比q＝-1/2则a6＝？

Answer 1

a2 + a5 = a1*q + a1*q^4 = a1*q * (1+q^3) = 18 (1)
a3 + a6 = a1*q^2 + a1*q^5 = a1*q^2 * (1+q^3) = 9 (2)
(2) / (1) q = 9 /18 = 1/2
把 q= 1/2 代入(1)得 a1*q * (1+q^3) = a1 * (1/2) * ( 1+ 1/8) =18 ==> a1 = 32
所以 通项式为
an = a1 * q^(n-1) = 32 *(1/2)^(n-1) = (2^5) *2^(1-n) = 2^(6-n)
a6=a1*q^5=32*(-1/2)^5=-1