The 20th term of an arithmetic sequence exceeds the 12th term by 64. If the units digit of the 20th term is 5, what is the units digit of the 2012th term? Solution a(20) = a + 19d a(12) = a + 11d (a+19d)-(a+11d) = 64 8d = 64 d = 64/8 = 8 a(20) = a + 19*8 = 152+a (mod 10) = 2+a (mod 10) = 5 a(2012) = a + 2011*8 (mod 10) = 5 + 1*8 (mod 10) = 13 mod 10 = 3 so the unit digit of 2012th term is 3..