Elements of Algebra

182 ELEMENTS SECT. III. And adding 2000 to both sides, we have $3000 \cdot (\frac{21}{20})^n = 1002000$. Then dividing both sides by 3000, we have $(\frac{21}{20})^n = 334$. Taking the logarithms, $n \log. \frac{21}{20} = \log. 334$; and dividing by $\log. \frac{21}{20}$, we obtain $n = \frac{\log. 334}{\log. \frac{21}{20}}$. Now, $\log. 334 = 2.5237465$, and $\log. \frac{21}{20} = 0.0211893$; therefore $n = \frac{2.5237465}{0.0211893}$; and, lastly, if we multiply the two terms of this fraction by 10000000, we shall have $n = \frac{25237465}{211893} = 119$ years, 1 month, 7 days; and this is the time, in which the principal of 1000l. will be increased to 1000000l. 554. But if we supposed that a person, instead of annually increasing his principal by a certain fixed sum, diminished it, by spending a certain sum every year, we should have the following gradations, as the values of that principal $a$, year after year, supposing it put out at 5 per cent, compound interest, and representing the sum which is annually taken from it by $b$: after 1 year, it would be $\frac{21}{20}a - b$, after 2 years, $(\frac{21}{20})^2 a - \frac{21}{20}b - b$, after 3 years, $(\frac{21}{20})^3 a - (\frac{21}{20})^2 b - \frac{21}{20}b - b$, after $n$ years, $(\frac{21}{20})^n a - (\frac{21}{20})^{n-1}b - (\frac{21}{20})^{n-2}b \dots - (\frac{21}{20})b - b$. 555. This principal consists of two parts, one of which is $(\frac{21}{20})^n \cdot a$, and the other, which must be subtracted from it, taking the terms inversely, forms the following geometrical progression: $b + (\frac{21}{20})b + (\frac{21}{20})^2 b + (\frac{21}{20})^3 b + \dots (\frac{21}{20})^{n-1}b$. Now we have already found (Art. 550.) that the sum of this progression is $20(\frac{21}{20})^n b - 20b$; if, therefore, we subtract this quantity from $(\frac{21}{20})^n \cdot a$, we shall have for the principal required, after $n$ years = $(\frac{21}{20})^n \cdot (a - 20b) + 20b$. 556. We might have deduced this formula immediately from that of Art. 550. For, in the same manner as we annually added the sum $b$, in the former supposition; so, in the present, we subtract the same sum $b$ every year. We have therefore only to put in the former formula, $-b$ every where, instead of $+b$. But it must here be particularly remarked, that if $20b$ is greater than $a$, the first part becomes negative, and, consequently, the principal will continually diminish. This will be easily perceived; for if we annually take away from the principal more than is added to it by the interest, it is evident that this principal must continually be-