设an=a1*q^(n-1), 有a2*a(n-1)=a1*an=128, 又a1+an=66, 知a1和an是方程x^2-66x+128=0的两根, 求得两根为2和64. 1)设a1=2,an=64, q^(n-1)=32, Sn=a1*(1-q^n)/(1-q)=a1*[1-q*q^(n-1)]/(1-q)=2*(1-32q)/(1-q)=126 得q=2...