How to Solve Common Terms in Two Arithmetic Sequences?

[UNknown 2010]

Homework Statement

Two finite arithmetic sequences :
1, 8, 15, 22, ... , 2003

2, 13, 24, 34, ... , 2004

HOW MANY TERMS THEY HAVE IN COMMON ??

I don't want the answer !

I want to know how to deal with this type of questions ?

Homework Equations

The Attempt at a Solution

 

[Mentallic]

Ok so the first sequence

$$T_{n1}=1,8,15,...,2003$$
$$T_{n1}=7n_1-6$$ for $$1\leq n_1\leq 287$$ for $$n_1\inZ$$

$$T_{n2}=2,13,24,...,2004$$

Find the pattern for this one too, then set the patterns equal and simplify to have $$n_1=f(n_2)$$ and so since n₁ is an integer you want fnnd when the function of n₂ is an integer for the specified integers n₂. You may need to find the pattern of the modulus (that is, if you know what modulus is, in other words, it's the remainder) for the first few integers n₂.

 

[HallsofIvy]

Those are both arithmetic series, the first having common difference 7, the second common difference 11.

The nth term of an arithmetic sequence with first term $a_0$ and common difference d is $a_n= a_0+ dn$

If, for example, you are asked what terms an arithmetic sequence with first term \(\displaystyle a_0\) and common difference d has in common with an arithmetic sequence with first term \(\displaystyle b_0\) and common difference e, you are asked to find number n and m such that \(\displaystyle a_0+ dn= b_0= em[/itex]. That can be reduced to $dn- em= b_0- a_0$ which is an example of a "Diophantine equation", a general equation of the form ax+ by= c for integers a, b, c and x,y.

To solve such a thing, first look for x and y such that ax+ by= 1. You can do that by using the "Euclidean division algorithm". For example, to solve 3x+ 8y= 1, note that 3 divides into 8 twice with remainder 2: 8= (2)3+ 2 or 8- (2)3= 2. Now, 2 divides into 3 once with remainder 1: 3= (1)2+ 1 or 3- (1)2= 1. Replacing the "2" in that equation by 8- (2)3 we have 3- (1)(8- (2)3)= (3)3- (1)8= (3)3+ (-1)(8)= 1. That is, x= 3 and y= -1 give a solution.

Notice that if we add any multiple of 8 to x and subtract that same multiple of 3 from y, we also get a solution: If 3x+ 8y= 1, then 3(x+ 8k)+ 8(y- 3k)= 3x+ 24k+ 8y- 24k= 3x+ 8y= 1 because the two "24k" terms cancel.

That is, for this problem, x= 3+ 8k and y= -1- 3k are integer solutions to 3x+ 8y= 1 for all integers k.

To solve, say 3x+ 8y= n for some n other than 1, just multiply x= 3+ 8k and y= -1- 3k by n.\)

 

[Dickfore]

Homework Statement

Two finite arithmetic sequences :
1, 8, 15, 22, ... , 2003

2, 13, 24, 34, ... , 2004

HOW MANY TERMS THEY HAVE IN COMMON ??

I don't want the answer !

I want to know how to deal with this type of questions ?

Homework Equations

The Attempt at a Solution

First of all, you need to find some regularity through the terms already given. One way to proceed is to calculate the subsequent differences. For the first one:
$$\begin{array}{ccc} 1 & & \\ & 7 & \\ 8 & & 0 \\ & 7 & \\ 15 & & 0 \\ & 7 & \\ 22 & & 0 \end{array}$$
In general, if the differences of order n are all constant, the sequence can be approximated by a polynomial of degree n. In this case $n = 1$. But, this is called arithmetic progression. The general term for this sequence is:

$$a_{n} = a_{1} + d (n - 1) = 1 + 7(n - 1) = 7n - 6$$

$$1 \le 7n - 6 \le 2003 \Leftrightarrow 7 \le 7n \le 2009 \Leftrightarrow 1 \le n \le 287$$

Can you do a similar analysis for the second sequence?