Elements of Algebra

132 ELEMENTS SECT. III. difference; so that when the first term and the difference are given, we may continue the arithmetical progression to any length. For example, if the first term be $2$, and the difference $3$, we shall have the following increasing progression: $2, 5, 8, 11, 14, 17, 20, 23, 26, 29$, &c. in which each term is found by adding the difference to the preceding term. 404. It is usual to write the natural numbers, $1, 2, 3, 4, 5$, &c. above the terms of such an arithmetical progression, in order that we may immediately perceive the rank which any 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. $1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9 \quad 10$ Arith. Prog. $2, 5, 8, 11, 14, 17, 20, 23, 26, 29$, &c. 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: $1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7$ $a, a+d, a+2d, a+3d, a+4d, a+5d, a+6d$, &c. according as the series is increasing, or decreasing; 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 $a + 99d$, and, generally, the nth term will be $a \pm (n-1)d$. 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 $a \pm (n-1)d$, according as the series is increasing or decreasing; which is consequently found by multiplying the difference by the number of terms minus one, and adding, or subtracting, that product from the first term. Suppose, for example, in an ascending arithmetical progression of a hundred terms, the first term is $4$, and the difference $3$; then the last term will be $99 \times 3 + 4 = 301$. 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 $z = a \pm (n-1)d$, 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 $n-1$.