在等比数列中,a5-a1=15,a4-a2=6,求a3和S8
问题描述：
在等比数列中,a5-a1=15,a4-a2=6,求a3和S8
答
an=a1q^(n-1)
a5-a1=15
a1(q^4-1) =15 (1)
a4-a2=6
a1q(q^2-1) =6 (2)
(2)/(1)
q/(q^2+1)=2/5
2q^2-5q+2 =0
(2q-1)(q-2)=0
q =1/2 or 2
case 1: q=1/2
a1= -16
a3= a1q^2 = -4
S8 = a1(q^8-1)/(q-1) = 16(255/256)/(1-1/2) = 255/8
case 1: q=2
a1=1
a3=a1q^2 =4
S8 = a1(q^8-1)/(q-1) = 255
答
∵{an}为等比数列∴a5-a1=a1(q^4-1)=15……①a4-a2=a1(q³-q)=6……②①/②式得：(q^4-1)/(q³-q)=15/6化简得：2q²-5q+2=0(q-2)(2q-1)=0解得：q=2或1/21°当q=2时,a1=15/(q^4-1)=1则a3=a1q²=1*2...