- #1

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## 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 ?

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- Thread starter UNknown 2010
- Start date

- #1

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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 ?

- #2

Mentallic

Homework Helper

- 3,798

- 94

Ok so the first sequence

[tex]T_{n1}=1,8,15,...,2003[/tex]

[tex]T_{n1}=7n_1-6[/tex] for [tex]1\leq n_1\leq 287[/tex] for [tex]n_1\inZ[/tex]

[tex]T_{n2}=2,13,24,...,2004[/tex]

Find the pattern for this one too, then set the patterns equal and simplify to have [tex]n_1=f(n_2)[/tex] and so since n

- #3

HallsofIvy

Science Advisor

Homework Helper

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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 [itex]a_0[/itex] and common difference d is [itex]a_n= a_0+ dn[/itex]

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 [itex]dn- em= b_0- a_0[/itex] which is an example of a "Diophantine equation", a general equation of the form ax+ by= c for

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.\)

- #4

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## 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:

[tex]

\begin{array}{ccc}

1 & & \\

& 7 & \\

8 & & 0 \\

& 7 & \\

15 & & 0 \\

& 7 & \\

22 & & 0

\end{array}

[/tex]

In general, if the differences of order

[tex]

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

[/tex]

[tex]

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

[/tex]

Can you do a similar analysis for the second sequence?

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