设公比是q
a5-a1
=a1*q^4-a1
=a1(q^4-1)
=a1(q^2+1)(q^2-1)=60 (i)
a4-a2
=a1*q^3-a1*q1
=a1q*(q^2-1)=24 (ii)
(i)/(ii)=>
(q^2+1)/q=60/24=5/2
=>2q^2-5q+2=0
=>(2q-1)(q-2)=0
=>q1=1/2,q2=2
因此这个数列的公比是1/2或2
设公比是q
a5-a1
=a1*q^4-a1
=a1(q^4-1)
=a1(q^2+1)(q^2-1)=60 (i)
a4-a2
=a1*q^3-a1*q1
=a1q*(q^2-1)=24 (ii)
(i)/(ii)=>
(q^2+1)/q=60/24=5/2
=>2q^2-5q+2=0
=>(2q-1)(q-2)=0
=>q1=1/2,q2=2
因此这个数列的公比是1/2或2