（a1+a1q+a1q^2+a1q^3+a1q^4）÷（a1q+a1q^2+a1q^3+a1q^4+a1q^5)=1/2既a1(1+q+q^2+q^3+q^4)÷a1q(1+q+q^2+q^3+q^4)=1/2既1/q=1/2既q=262-31=a1q^5-a1=31所以a1×2^5-a1=3131a1=31所以a1=1所以通项公式为an=2^(n-1)谢谢，你讲的很清楚那就采纳吧