Question

等比数列{an}中，a1+a3=10,a4+a6=5/4,则数列{an}的通项公式为？

Answer 1

a1+a3=10,
a4+a6=a1×q³+a3×q³=(a1+a3)q³=5/4
(a4+a6)/(a1+a3)=(a1+a3)q³/(a1+a3)=q³=5/4÷10=1/8
q=1/2
a1+a3=a1(1+q²)=a1×(1+(1/2)²)=10
a1=8
an=a1×q^(n-1)=8×(1/2)^(n-1)=2^(4-n)
数列{an}的通项公式:an=2^(4-n)

Answer 2

a1+a3=10,a4+a6=5/4,
a1+a1q^2=10
a1(1+q^2)=10
a1q^3+a1q^5=5/4
a1q^3(1+q^2)=5/4
10aq^3=5/4
q^3=1/8
q=1/2
a1(1+1/4)=10
a1=8
an=a1q^(n-1)=16(1/2)^n=2^(4-n)

Answer 3

a4+a6=a1q3+a3q3=(a1+a3)q3=5/4,
q3=1/8,q=1/2

a1+a3=a1(1+q2)=10,
a1=8

an=8*(1/2)的（n-1)次
=(1/2)的（n-4)次

Answer 4

设数列{an}的公比为q，则（a4+a6）/（a1+a3）=q^3=1/8,故q=1/2，则a1+a3=a1+a1q^2=（5/4）a1=10，故a1=8，故an=2^（4-n）。

Answer 5

a1 + a1*q^2 = 10a1*q^3 + a1*q^5 =q^3*（a1 + a1*q^2）= 5/4解得：a1 = 8 , q = 1/2an = a1*q^(n-1) = 2^(n+2)

Answer 6

通项公式an＝2的（4－n）次方