Given two arithmetic sequences...

carnage · #1 $Old$ 05-22-2008, 02:10 PM

Given the following 2 arithmetic sequences, determine the first ten terms common to both sequences.

Sequence #1
3, 14, 25, ...

Sequence #2
2, 9, 16, ...

I figured out that the two general formulas would be tn = 3 + 11 (n-1) and tn = 2 + 7 (n-1), respectively... But I'm not sure where to go from there. I've asked for help elsewhere but I still don't get it, and my test is coming up soon... =/ Thanks a lot.

carnage · #2 $Old$ 05-22-2008, 02:17 PM

Quote:

Originally Posted by galactus $View Post$

They just want the first ten terms, just add 11 in the first and 7 in the second.

The first one: 3,14,25,36,47,58,69,....... and so on

Is there no other way to do it besides writing the entire sequences out, and then circling the ones in common?

Mathnasium · #3 $Old$ 05-22-2008, 03:14 PM

If I'm reading your original post correctly, you don't want the first ten terms of each sequence, right? You want the first ten terms that are COMMON to BOTH sequences. If so:

Here's what I did - it's slightly better than listing out every term.

The first sequence's terms can be described by 2 + 7x.

The second sequence's terms can be described by 3 + 11y.

I want to know when these are equal, so:

2 + 7x = 3 + 11y.

I solved for y and got:

$y = \frac{7}{11}x - \frac{1}{11}$ .

So I'm looking for an integer value of x that gives me an integer value for 7 (since all the numbers in both sequences are integers). I found the first one by trial and error. x=8 gives y=5. So the point (8,5) is on my line, and gives 58 for both 2 + 7x and 3 + 11y. (Points in which both coordinates are integers are called "lattice points" - there may be an easier way to find them, but I don't know it. Essentially, that's what we're looking for here - lattice points on our line.)

So 58 is in both sequences. Now, noting that the slope of our line is 7/11, we can increase our y value by 7 and our x value by 11 (since slope is change in y / change in x) to get the next value of x and y that gives integer coordinates: (19, 12). Plugging x = 19 and y = 12 into the appropriate equations gives 135 in both, so that's the next term.

Keep using slope to "jump" from each integer-coordinate point to the next. Use x and y, plug them into your equations, and you should be good to go:

(8,5), (19,12), (30, 19), (41, 26), ...

Hope this makes sense. I kind of figured it out while writing this, so it's a little... disorganized.

Mathstud28 · #4 $Old$ 05-22-2008, 03:17 PM

Quote:

Originally Posted by Mathnasium $View Post$

If I'm reading your original post correctly, you don't want the first ten terms of each sequence, right? You want the first ten terms that are COMMON to BOTH sequences. If so:

Here's what I did - it's slightly better than listing out every term.

The first sequence's terms can be described by 2 + 7x.

The second sequence's terms can be described by 3 + 11y.

I want to know when these are equal, so:

2 + 7x = 3 + 11y.

I solved for y and got:

$y = \frac{7}{11}x - \frac{1}{11}$ .

So I'm looking for an integer value of x that gives me an integer value for 7 (since all the numbers in both sequences are integers). I found the first one by trial and error. x=8 gives y=5. So the point (8,5) is on my line, and gives 58 for both 2 + 7x and 3 + 11y.

So 58 is in both sequences. Now, noting that the slope of our line is 7/11, we can increase our y value by 7 and our x value by 11 (since slope is change in y / change in x) to get the next value of x and y that gives integer coordinates: (19, 12). Plugging x = 19 and y = 12 into the appropriate equations gives 135 in both, so that's the next term.

Keep using slope to "jump" from each integer-coordinate point to the next. Use x and y, plug them into your equations, and you should be good to go:

(8,5), (19,12), (30, 19), (41, 26), ...

Hope this makes sense. I kind of figured it out while writing this, so it's a little... disorganized.

Could you not use the variation of the Euclidean algorithim on this since it is a Diophantine equation?

Or am I stepping out of my bounds here?

Plato · #5 $Old$ 05-22-2008, 05:03 PM

The answer that I posted before was correct, BUT the notation was mistaken.
Allow me to change the notation to an index of zero:
$a_0 = 2,\,a_1 = 9,\, \cdots ,\,a_n = 2 + 7n$ & $b_0 = 3,\,b_1 = 14,\, \cdots ,\,b_n = 3 + 11n$ .

Now it is easy to check that: $a_8 = b_5 ,\,a_{19} = b_{12} ,\,a_{30} = b_{19} ,\, \cdots$ .

Do you see the relation among the subscripts: 8,19,30 & 5,12,19?
Are they $- 3 + 11j,\,j = 1,2,3, \cdots$ and $- 2 + 7k,\,k = 1,2,3, \cdots$ ?