[MATH] ARITHMETIC SEQUENCES

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

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