设等差数列an公差为d,等比数列bn公比为q
a1=b1
a3=b3 => a1+2d=b3=a1*q^2
a7=b5 => a1+6d=b5=a1*q^4
b3²=b1*b5 => (a1+2d)²=a1*(a1+6d) =>a1=2d
所以b3=a3=4d,b5=a7=8d
因为bn是各项为正的等比数列,所以q=√(b5/b3)=√2
bn=b1*q^(n-1）=2d*√2^(n-1)=2d*2^[(n-1)/2]
a15=a1+14d=16d=2d*8=2d*2^3
bm=2d*2^[(m-1)/2]
由a15=bm得(m-1)/2=3
m=7