Elements of Algebra

CHAP. III. OF ALGEBRA. 133 in order to obtain the required value of the difference $d$, which will be $\frac{z-a}{n-1}$. 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, $1\frac{1}{9}$, $1\frac{2}{9}$, $1\frac{3}{9}$, $1\frac{4}{9}$, $1\frac{5}{9}$, $1\frac{6}{9}$, $1\frac{7}{9}$, $1\frac{8}{9}$, 2. Another example. Let the first term be $2\frac{1}{2}$, the last term $12\frac{1}{2}$, and the number of terms 7; the difference will be $\frac{12\frac{1}{2}-2\frac{1}{2}}{7-1} = \frac{10}{6} = \frac{5}{3} = 1\frac{2}{3}$, and consequently the progression: 1 2 3 4 5 6 7 $2\frac{1}{2}$, $4\frac{1}{6}$, $5\frac{5}{6}$, $7\frac{1}{2}$, $9\frac{1}{6}$, $10\frac{5}{6}$, $12\frac{1}{2}$. 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 $\frac{z-a}{d} = n-1$; 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 $\frac{100-4}{12} + 1 = 9$;