a5=a3+2d
d=2,a1=-2
a5*a5=a3*ak1
ak1=18
公比q=3
此时akn=ak1*q^(n-1)=18*3^(n-1)
akn是{an}的项
akn=a1+(kn-1)d=-2+2(kn-1)=2kn-4
akn=2kn-4=18*3^(n-1)
得到kn=9*3^(n-1)+2
设数列{an}是等差数列,a5=6
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