设an是等差数列,bn是各项都为正数的等比数列.a1=b1=1,a3+b5=21,a5+b3=13求an,bn的通项公式

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设公差为d,公比为q,
a3+b5=21,a5+b3=13
a1+2d+b1*q^4=21
a1+4d+b1q²=13
即
1+2d+q^4=21
1+4d+q²=13
解得
2q^4-q²-28=0
（2q²+7）（q²-4）=0
因为q>0
所以
d=2
q=2
所以
an=a1+(n-1)d=1+2(n-1)
即
an=2n-1
bn=b1*q^(n-1)=2^(n-1)
即
bn=2^(n-1)

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