设{an}等差数列公差为d,{bn}等比数列公比为q（q＞0）
a3+b5=1+2d+1×q^4=19,2d+q^4=18①
a5+b3=1+4d+1×q²=9,4d+q²=8②
①×2-②得：2q^4-q²=28,解得q²=4（-7/2舍去）q=2,d=1
{an}的通项公式an=n
{bn}的通项公式bn=2^(n-1)
{an}为等差数列,{bn}为各项都是正数的等比数列,且a1=b1=1,a3+b5=19,a5+b3=9,求{an}和{
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