在等比数列{an}中，

Question

在等比数列{an}中，
（1）a4=27，q= - 3，求a7；
（2）a2=18，a4=8，求a1与q；
（3）a5=4，a7=6，求a9；
（4）a5 - a1=15，a4 - a2=6，求a3.
请写出详细的解答过程，谢谢。

Answer 1

（1）a4=27，q= - 3，求a7； a7= a4 * [q^(7-4)] = 27*[(-3)^3]=27*(-27)= -729 所以 a7 = -729。
（2）a2=18，a4=8，求a1与q； a2=a1 * q =18 a4=a1 * q^3 = 8 a4/a2 = (a1*q^3)/(a1*q) = q^2 =8/18=4/9 ==> q= ±2/3 a2=a1 * q =18 ==> a1 = 18/q = 18/(±2/3) = ±18*3/2 = ±27 所以 a1 = ±27，q= ±2/3。
（3）a5=4，a7=6，求a9； a5 = a1 * q^4 = 4 a7 = a1 * q^6 = 6 a7/a5 = q^2 = 6/4 = 3/2 ==> q^2= 3/2 a1 = 4 / (q^4) = 4 / [(3/2)^2] = 4 / (9/4) = 16/9 a9 = a1 * q^8 = (16/9) * [(3/2)^4] = (16/9) * (81/16) = 9
（4）a5 - a1=15，a4 - a2=6，求a3 a5 - a1 = 15 ==> a1*(q^4 -1) =15 a4 - a2 = 6 ==> a1 * (q^3 - q) = 6 上面两式相除，则有 (q^4 -1) / (q^3 - q) = 15/6 (q≠0, ±1)==> [(q^2 + 1) * (q^2 - 1)] / [q * (q^2 - 1)] = 5/2==> (q^2 + 1) / q =5/2==> 2*(q^2) + 2 = 5*q ==> 2*(q^2) - 5*q + 2 = 0==> (2q-1)*(q-2)=0==> q= 1/2 或 q= 2 当 q = 1/2 时， a1 = 6 / (q^3 - q) = 6 / [(1/2)^3 - 1/2] = 6 / (-3/8) = -16 a3 = a1 * q^2 = -16 * (1/2)^2 = -16 * (1/4) = -4 当 q = 2 时， a1 = 6 / (q^3 - q) = 6 / (2^3 - 2) = 6/6 = 1 a3 = a1 * q^2 = 1 * 2^2 = 1 * 4 = 4。

Answer 2

1.a7=a4×q^3=27×(-3)^3=-7292.因为a4/a2=q^2所以q^2=8/18=4/9q=2/3或-2/3a1=a2/q=18/(2/3)=27或18/(-2/3)=-273.因为等比数列中，a7是等比数列所以(a7)^2=a5a96^2-4a9a9=94.因为a5-a1=a3q^2-a3/q^2=15a4-a2=a3q-a3/q=6所以a5-a1/a4-a2=5/2=(q^2- 1/q^2)/(q -1/q)=q+ 1/q所以q=2或1/2因为a4-a2=a3(q- 1/q)=6当q=2时，a3=6/(3/2)=4当q=1/2时，a3=6/(-3/2)=-4