∵a1∥a2,a2⊥a3,∴a1⊥a3,又∵a3∥a4,∴a1⊥a4,又∵a4⊥a5,∴a1∥a5,又∵a5∥a6,∴a1∥a6,又∵a6⊥a7,∴a1⊥a7,又∵a7∥a8,∴a1⊥a8,又∵a8⊥a9,∴a1∥a9,又∵a9∥a10,∴a1∥a10.由以上可得规律：a1⊥a(4n+3),a1⊥a...