设公比qa5q^2=q5q+2q5即q^2-q-2=0解得q=2或q=-1(不合）又根号下am*an等于2倍根号2a1,即根号下am*an等于(根号8)a1则am*an=8a1a1*2^(m-1)*a1*2^(n-1)=8a1^22^(m+n-2)=82^(m+n)=32得m+n=5因mn为正整数,所以当m=2,n=3时...