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How much different arithmetic sequences can you make from the numbers 1 to 51 ? Note: The sequence length has to be 3 numbers. The difference between each 2 numbers is positive.

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    $\begingroup$ To clarify the problem, perhaps you could give a few examples and a few non-examples of allowed sequences? $\endgroup$ – Chris Culter Aug 16 '13 at 23:54
  • $\begingroup$ Both answers assume you mean the sequences to be arithmetic progression. Do you? Is $1,3,9$ a valid sequence? It is a different problem, also interesting, if it is. Can you use these ideas to solve that one? $\endgroup$ – Ross Millikan Aug 17 '13 at 0:20
  • $\begingroup$ It has to be an arithmetic progression with 3 numbers and the difference has to be positive, ex. : 1,2,3 1,26,51 etc... $\endgroup$ – HaloKiller Aug 17 '13 at 0:29
  • $\begingroup$ @YonatanMarkman And once you learn the approaches, you can try this similar problem. $\endgroup$ – Calvin Lin Aug 17 '13 at 0:40
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A valid arithmetic sequence of length $3$ and of constant (integer) difference $d>0$ consists of the numbers $$ n,n+d,n+2d $$ For some integer $n$. We therefore can state that we will have a valid sequence as long as, given $n$ from $1$ to $51$, $n+2d\leq 51$.

That is, for any $n$, we have a valid sequence if we choose a $d$ such that $2d\leq 51-n$.

For $51,50$, there is no valid choice of $d$. For $49,48$, there is one valid choice of $d$. Keep going down in this fashion and you find that the pattern continues so that with $n=3,2$, there are $24$ valid choices, and for $n=1$, there are $25$.

Add all these up to get the total number and you have $N$, the number of valid sequences, is $$ \begin{align} N &= 1+1+2+2+3+3+\cdots+24+24+25 \\ &= 2\times(1+2+\cdots+24)+25\\ &= 2\frac{24\times25}{2}+25\\ &= 25\times25\\ &= 625 \end{align} $$

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There are several ways to count in an organized manner.. We describe three of them. (Our favourite way is the first, and relatives.)

First way: We need to choose two numbers $a$ and $b$ to serve as the ends of our sequence. The numbers $a$ and $b$ determine a three-term increasing arithmetic sequence precisely if $\frac{a+b}{2}$ is an integer. So we want $a$ and $b$ to have the same parity.

There are $25$ even numbers in our interval. We can choose $2$ of them in $\binom{25}{2}$ ways. There are $26$ odd numbers in the interval. We can choose $2$ of them in $\binom{26}{2}$ ways.

Thus the total is $\binom{25}{2}+\binom{26}{2}$.

Second way: There is one three-term increasing arithmetic sequence with "middle" number $2$, there are $2$ with middle number $3$, there are $3$ with middle number $4$, and so on up to $24$ with middle number $25$.

There are $25$ with middle number $26$.

And by symmetry there are just as many with middle number $\gt 26$ as there are with middle number $\lt 26$.

Thus our count is $2(1/2)(24)(25)+25$, that is, $25^2$.

Third way: There is $1$ (three-term increasing) arithmetic sequence with common difference $25$, there are $3$ with common difference $24$, there are $5$ with common difference $23$, and so on up to $49$ with common difference $1$.

The sum $1+3+5+\cdots+49$ of the odd numbers up to $49$, is, by a standard formula, equal to $25^2$.

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  • $\begingroup$ +1 I think you should present your favorite way first ... $\endgroup$ – Calvin Lin Aug 17 '13 at 0:40

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