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

Algebra ->  Sequences-and-series -> 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      Log On


   

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) About Me  (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.