標題:
[MATH] ARITHMETIC SEQUENCES
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發問:
Consider two different arthimetic sequences: 3,14,25,36,47,58,69,80... 2,9,16,23,30,37,44,51... What are the first 10 numbers that appear in both sequences?
最佳解答:
For the sequence 3,14,25,36,47,58,69,80... First term =3, difference d1=11 So the m th term is 3+(m-1)(11)=11m-8 For the sequence 2,9,16,23,30,37,44,51... First term =2, difference d1=7 So the n th term is 2+(n-1)(7)=7n-5 If a number appear in both sequence, then 11m-8=7n-5 11m=7n+3 Consider 11m-7n=1 Then m=2, n=3 is one solution So the general solution of 11m-7n=1 is m=2+7t, n=3+11t (where t is an integer) And then the general solution of 11m-7n=3 is m=6+21t, n=9+33t (where t is an integer) substitute t=0,1,2...9. We have the table as follows t m n value 0 6 9 58 1 27 42 289 2 48 75 520 3 69 108 751 4 90 141 982 5 111 174 1213 6 132 207 1444 7 153 240 1675 8 174 273 1906 9 195 306 2137 So, the first 10 numbers that appear in both sequences are 58,289,520,751,982,1213,1444,1675,1906,2137
其他解答:
a = 3, d = 11 T(n) = 3 + (n-1) * 11 T(1) 至 T(7) 寫左出黎 T(8) = 80, T(9) = 91, T(10) = 102 a = 2, d = 7 S(n) = 2 + (n-1) * 7 S(1) 至 S(7) 寫左出黎 S(8) = 51, S(9) = 58, S(10) = 65
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