Elements of Algebra

134 ELEMENTS SECT. III. 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 $1\frac{1}{3}$, 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, $3\frac{1}{3}$, $4\frac{2}{3}$, 6. Again, let the first term be $3\frac{1}{2}$, the last $7\frac{1}{2}$, and the difference $1\frac{1}{3}$, the number of terms will be $\frac{7\frac{1}{2}-3\frac{1}{2}}{1\frac{1}{3}} + 1 = 4$; which are, $3\frac{1}{2}$, $4\frac{5}{6}$, $6\frac{1}{6}$, $7\frac{1}{2}$. 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 : 1st. The first term, $a$; 2d. The last term, $z$; 3d. The difference, $d$; and 4th. 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 $d = \frac{z-a}{n-1}$; and 4. If $a$, $z$, and $d$, are known, we have $n = \frac{z-a}{d} + 1$.