### Part I. Section III. Chapter 3. “Of Arithmetical Progressions.”

402 We have already remarked, that a series of numbers composed of any number of terms, which always increase, or decrease, by the same quantity, is called an arithmetical progression.

Thus, the natural numbers written in their order, as

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, etc.

form an arithmetical progression, because they constantly increase by unity; and the series

25, 22, 19, 16, 13, 10, 7, 4, 1, etc.

is also such a progression, since the numbers constantly decrease by 3.

403 The number, or quantity, by which the terms of an arithmetical progression become greater or less, is called the difference; so that when the first term and the difference are given, we may continue the arithmetical progression to any length.

For example, let the first term be 2, and the difference 3, and we shall have the following increasing progression:

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, etc.

in which each term is found by adding the difference to the preceding one.

404 It is usual to write the natural numbers, 1, 2, 3, 4, 5, etc. above the terms of such an arithmetical progression, in order that we may immediately perceive the rank, which anv term holds in the progression, which numbers, when written above the terms, are called indices; thus, the above example will be written as follows:

 Indices 1 2 3 4 5 6 7 8 9 10 etc. Arithmetic Progression 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, etc.

where we see that 29 is the tenth term.

405 Let $$a$$ be the first term, and $$d$$ the difference, the arithmetical progression will go on in the following order:

 1 2 3 4 5 6 7 8 etc. $$a$$, $$a+d$$, $$a+2d$$, $$a+3d$$, $$a+4d$$, $$a+5d$$, $$a+6d$$, $$a+7d$$, etc.

whence it appears that any term of the progression might be easily found, without the necessity of finding all the preceding ones, by means only of the first term $$a$$ and the difference $$d$$; thus, for example, the tenth term will be $$a + 9d$$, the hundredth term $$a+9d$$, and, generally, the $n$th term will be

$a + (n -1)d.$

406 When we stop at any point of the progression, it is of importance to attend to the first and the last term, since the index of the last term will represent the number of terms. If, therefore, the first term be $$a$$, the difference $$d$$, and the number of terms $$n$$, we shall have for the last term $$a + (n -1)d$$, which is consequently found by multiplying the difference by the number of terms minus one, and adding that product to the first term. Suppose, for example, in an arithmetical progression of a hundred terms, the first term is 4, and the difference 3; then the last term will be 99·3 + 4 = 301.

407 When we know the first term $$a$$, and the last $$z$$, with the number of terms $$n$$, we can find the difference $$d$$; for, since the last term

$z = a + (n -1)d,$

if we subtract a from both sides, we obtain $$z - a = (n-1)d$$. So that by taking the difference between the first and last term, we have the product of the difference multiplied by the number of terms minus 1; we have therefore only to divide $$z - a$$ by $$n - 1$$ in order to obtain the required value of the difference $$d$$, which will be $$\dfrac{z-a}{n-1}$$. This result furnishes the following rule: Subtract the first term from the last, divide the remainder by the number of terms minus 1, and the quotient will be the common difference: by means of which we may write the whole progression.

408 Suppose, for example, that we have an increasing arithmetical progression of nine terms, whose first is 2 and last 26, and that it is required to find the difference. We must subtract the first term 2 from the last 26, and divide the remainder, which is 24, by 9 - 1, that is, by 8; the quotient 3 will be equal to the difference required, and the whole progression will be:

 1 2 3 4 5 6 7 8 9 2, 5, 8, 11, 14, 17, 20, 23, 26

To give another example, let us suppose that the first term is 1, the last 2, the number of terms 10, and that the arithmetical progression, answering to these suppositions, is required; we shall immediately have for the difference $$\frac{2-1}{10-1}=\frac{1}{9}$$, and thence conclude that the progression is:

 1 2 3 4 5 6 7 8 9 10 1, 1⅑, 1²⁄₉, 1³⁄₉, 1⁴⁄₉, 1⁵⁄₉, 1⁶⁄₉, 1⁷⁄₉, 1⁸⁄₉, 2

Another example. Let the first term be 2⅓, the last term 12½, and the number of terms 7; the difference will be

$\frac{12\frac{1}{2}-2\frac{1}{3}}{7-1}=\frac{10\frac{1}{6}}{6} = \frac{61}{36} = 1\frac{25}{36},$

and consequently the progression:

 1 2 3 4 5 6 7 2⅓, 4¹⁄₃₆, 5¹³⁄₁₈, 7⁵⁄₁₂, 9⅑, 10²⁹⁄₃₆, 12½.

409 If now the first term $$a$$, the last term $$z$$, and the difference $$d$$, are given, we may from them find the number of terms $$n$$; for since $$z - a = (n - 1)d$$, by dividing both sides by $$d$$, we have $$n-1 = \frac{z-a}{d}$$; also $$n$$ being greater by 1 than $$n - 1$$, we have

$n = \frac{z-a}{d}+1;$

consequently the number of terms is found by dividing the difference between the first and the last term, or $$z -a$$, by the difference of the progression, and adding unity to the quotient.

For example, let the first term be 4, the last 100, and the difference 12, the number of terms will be $$\frac{100-4}{12}+1=9$$; and these nine terms will be,

 1 2 3 4 5 6 7 8 9 4, 16, 28, 40, 52, 64, 76, 88, 100

If the first term be 2, the last 6, and the difference 1⅓, the number of terms will be $$\frac{4}{1\frac{1}{3}}+1=4$$; and these four terms will be,

 1 2 3 4 2, 3⅓, 4⅔, 6

Again, let the first term be 3⅓, the last 7⅔, and the difference 1⁴⁄₉, the number of terms will be $$\frac{7\frac{2}{3}-3\frac{1}{3}}{1\frac{4}{9}}+1=4$$; which are,

3⅓, 4⁷⁄₉, 6²⁄₉, 7⅔.

410 It must be observed, however, that as the number of terms is necessarily an integer, if we had not obtained such a number for $$n$$, in the examples of the preceding article, the questions would have been absurd.

Whenever we do not obtain an integer number for the value of $$\frac{z-a}{d}$$, it will be impossible to resolve the question; and consequently, in order that questions of this kind may be possible, $$z - a$$ must be divisible by $$d$$.

411 From what has been said, it may be concluded, that we have always four quantities, or things, to consider in an arithmetical progression:

1. The first term, $$a$$;
2. The last term, $$z$$;
3. The difference, $$d$$; and
4. The number of terms, $$n$$.

The relations of these quantities to each other are such, that if we know three of them, we are able to determine the fourth; for,

1. If $$a$$, $$d$$, and $$n$$, are known, we have $$z=a+(n-1)d$$.
2. If $$z$$, $$d$$, and $$n$$, are known, we have $$a = z-(n-1)d$$.
3. If $$a$$, $$z$$, and $$n$$, are known, we have $$d = \frac{z-a}{n-1}$$.
4. If $$a$$, $$z$$, and $$d$$, are known, we have $$n=\frac{z-a}{d}=1$$.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.