已知等差数列{an}和等比数列{bn}满足:a1+b1=3,a2+b2=7,a3+b3=15,a4+b4=35,则a5+b5=______

∵a1+b1=3，①a2+b2=a1+d+b1q=7，②a3+b3=a1+2d+b1q2=15，③a4+b4=a1+3d+b1q3=35④②-①可得，4-d=b1（q-1）③-②可得，8-d=b1q（q-1）④-③可得，20-d=b1q2(q-1)∴4-d8-d=1q，8-d20-d=1q∴4-d8-d=8-d20-d解方程可求d=2，q=3，b1=1，a1=2∴a5+b5=10+81=91故答案为：91

数列｛an｝，｛bn｝都是等差数列
设数列｛an｝公差是da，｛bn｝公差是db
a1+b1=7
a3+b3=a1+2da+b1+2db=7+2(da+db)=21
da+db=7

a5+b5
=a1+4da+b1+4db
=a1+b1+4(da+db)
=7+4*7
=35