# Quod Erat Demonstrandum

## 2012/05/29

### an easy question about A.S.

Filed under: NSS — johnmayhk @ 12:45 下午
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Many students think that ‘arithmetic sequences’（等差數列）is an extremely easy topic in core mathematics, however, I guess many of them may have misunderstanding when they tackle certain problems like:

Let $T_n=7n+1$, $U_n=11n-4$（where n is any positive integer）

How many terms of sequences $\{T_n\}$ and $\{U_n\}$ are in common for $1\le n \le 2012$?

This is just a 1-mark question I’d set in a quiz.

Many students thought that

$7n+4$ and $11n-1$ could be regarded as two straight lines, because they are not parallel to each other, there should be an intersection, and hence there should be only one term in common."

No!

Undoubtedly, there is exactly one intersection between the lines

$y=7x+4$
$y=11x-1$

however, to find common terms of two sequences,

$y=7n+4$
$y=11n-1$

we are not looking for the intersection of lines, instead, we are searching for the points having same y-coordinates; where their n-coordinates (or x-coordinates) may not be the same.

A picture is worth a thousand words, let’s refer to the following

As for example, when $n=15$, $T_n=7(15)+1=106$; when $n=10$, $U_n=11(10)-4=106$.

Hence, 106 is a common term of sequences $\{T_n\}$ and $\{U_n\}$.

Now, how to solve the original question?

Many.

May be, first of all, we find out one common term, like 106.

That is

$7(15)+1=11(10)-4$

So, for any integer $k$, we have

$7(15+11k)+1=11(10+7k)-4$

(note that 7 and 11 are coprime)

For requiring $1\le n \le 2012$, just set

($15+11k \ge 1$ and $10+7k \ge 1$) and ($15+11k \le 2012$ and $10+7k \le 2012$)

yield,

$-1\le k \le 181$,

it means that there are 183 terms in common, namely

29,106,183,…,13966,14043.