SOLUTION: Given two different arithmetic sequences 3,14,25,36 (obviously increasing by 11) and 2,9,16,23 (obviously increasing by 7). What are the first 10 numbers which appear in both seq

Question 859811: Given two different arithmetic sequences 3,14,25,36 (obviously increasing by 11)
and 2,9,16,23 (obviously increasing by 7). What are the first 10 numbers which appear in both sequences ?
One tutor answered it with a solution that I had already done (the long way) which is simply to "keep adding 11 and 7 until you have 10 terms that match. All of which is true, but isn't there a formula that can be used to solve it a different way?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
An = A1 + (n - 1)d
An=3 + 11(n - 1)
An=2 + 7(n - 1)
You might think that by setting them equal you would get at least the first matching number but you don't.
3 + 11(n - 1)= 2 + 7(n - 1)
and solve for n
3-11n-11=2+7n-7
11n-8=7n-5
4n=3
n=3/4
The problem seems to be that the matching numbers don't occur at the same place in the sequence so n is not the same for both. They never coincide in the same place.

3 + 11(x - 1)= 2 + 7(y - 1)
The solution pattern is
x = 7a+6 and y = 11a+9
now just pick your a and solve for x and y
the first solution with a=1 is
x = 13 and y = 20
Ax=3 + 11(x - 1)
Ay=2 + 7(y - 1)
Ax=Ay
and then plug those into the formula.
Ax=3 + 11(x - 1):3+11*12=135
Ay=2 + 7(y - 1) : 2+7*19=135
when a=2
x = 7a+6 and y = 11a+9
x=20 y=31
Ax=3 + 11(x - 1):3+11*19=212
Ay=2 + 7(y - 1)=2+210=212
and so on till you have your ten matches.

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