Elements of Algebra

CHAP. XI. OF ALGEBRA. 169 Consequently, $s = \frac{a}{1 - \frac{b}{c}} = \frac{ac}{c-b}$, by multiplying both the numerator and denominator by $c$. The sum of the infinite geometrical progression proposed is, therefore, found by dividing the first term $a$ by $1$ minus the ratio; or by multiplying the first term $a$ by the denominator of the ratio, and dividing the product by the same denominator diminished by the numerator of the ratio. 521. In the same manner we find the sums of progressions, the terms of which are alternately affected by the signs $+$ and $-$. Suppose, for example, $s = a - \frac{ab}{c} + \frac{ab^2}{c^2} - \frac{ab^3}{c^3} + \frac{ab^4}{c^4} - \ldots$ Multiplying by $\frac{b}{c}$, we have, $\frac{b}{c}s = \frac{ab}{c} - \frac{ab^2}{c^2} + \frac{ab^3}{c^3} - \frac{ab^4}{c^4} + \ldots$ And, adding this equation to the preceding, we obtain $(1 + \frac{b}{c})s = a$: whence we deduce the sum required, $s = \frac{a}{1 + \frac{b}{c}}$, or $s = \frac{ac}{c+b}$. 522. It is evident, therefore, that if the first term $a = \frac{3}{5}$, and the ratio be $\frac{2}{5}$, that is to say, $b=2$, and $c=5$, we shall find the sum of the progression $\frac{3}{5} + \frac{6}{25} + \frac{12}{125} + \frac{24}{625} + \ldots = 1$; since, by subtracting the ratio from $1$, there remains $\frac{3}{5}$, and by dividing the first term by that remainder, the quotient is $1$. It is also evident, if the terms be alternately positive and negative, and the progression assume this form: $\frac{3}{5} - \frac{6}{25} + \frac{12}{125} - \frac{24}{625} + \ldots$ that the sum will be $\frac{a}{1 + \frac{b}{c}} = \frac{\frac{3}{5}}{1 + \frac{2}{5}} = \frac{3}{7}$. 523. Again: let there be proposed the infinite progression, $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \frac{3}{10000} + \frac{3}{100000} + \ldots$ The first term is here $\frac{3}{10}$, and the ratio is $\frac{1}{10}$; therefore