Question

在等比数列中,a5-a1=15,a4-a2=6,求a3和S8

Answer 1

∵{an}为等比数列
∴a5-a1=a1(q^4-1)=15……①
a4-a2=a1(q³-q)=6……②
①/②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1/2
1°当q=2时，a1=15/(q^4-1)=1
则a3=a1q²=1*2²=4
S8=a1(1-q^8)/(1-q)=1*(1-2^8)/(1-2)=255
2°当q=1/2时，a1=15/(q^4-1)=-16
则a3=a1q²=-16*(1/2)²=-1
S8=a1(1-q^8)/(1-q)=-16*[1-(1/2)^8]/(1-1/2)=-255/8

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Answer 2

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