Question

求解一道高一数学题，请写出详细过程，谢谢了。

在等比数列A n中，a9+a10=a,(a不等于0），a19+a20=b,则a99+a100=?

Answer 1

a9 = a1*q^8
a10 = a1*q^9
a9+a10 = a1*q^8(1+q) = a
a19 + a20 = a1*q^18(1+q) = b
q^10 = b/a
a1*(1+q ) = a/ q^8 = b/q^18
a99 + a100 = a1*q^98 *(1+q) = a / q^8 * q^98 = a* q^90 = a *(b/a)^9 = b^9 / a^8

Answer 2

设公比为q，
a9+a10=a1q^8+a1q^9=a1q^8(1+q)=a
同理，a19+a20=a1q^18(1+q)=b，
两式相除，得q^10=b/a
a99+a10=a1q^98(1+q)=a1q^8(1+q)q^90=a*(b/a)^9=b^9/a^8

Answer 3

a19+a20=a9*(q^10)+a10*(q^10)=(q^10)*(a9+a10)=b

a9+a10=a

所以q^10=b/a

a99+a100
=a9*(q^90)+a10*(q^90)
=a*(q^90)
=a*((b/a)^9)
=(b^9)/(a^8)