Elements of Algebra

CHAP. III. OF ALGEBRA. 131 same quantity; as in the numbers $4, 7, 10$; and the decreasing progression is that in which the terms go on always diminishing by the same quantity, such as the numbers $9, 5, 1$. 399. Let us suppose the numbers $a, b, c$, to be in arithmetical progression; then $a - b = b - c$, whence it follows, from the equality between the sum of the extremes and that of the means, that $2b = a + c$; and if we subtract $a$ from both, we have $2b - a = c$. 400. So that when the first two terms $a, b$, of an arithmetical progression are given, the third is found by taking the first from twice the second. Let $1$ and $3$ be the first two terms of an arithmetical progression, the third will then be $2 \times 3 - 1 = 5$; and these three numbers $1, 3, 5$, give the proportion $1 - 3 = 3 - 5$. 401. By following the same method, we may pursue the arithmetical progression as far as we please; we have only to find the fourth term by means of the second and third, in the same manner as we determined the third by means of the first and second, and so on. Let $a$ be the first term, and $b$ the second, the third will be $2b - a$, the fourth $4b - 2a - b = 3b - 2a$, the fifth $6b - 4a - 2b + a = 4b - 3a$, the sixth $8b - 6a - 3b + 2a = 5b - 4a$, the seventh $10b - 8a - 4b + 3a = 6b - 5a$, &c. CHAP. III. 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 $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \&c.$ form an arithmetical progression, because they constantly increase by unity; and the series $25, 22, 19, 16, 13, 10, 7, 4, 1, \&c.$ 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 K 2