Elements of Algebra

CHAP. XIII. OF ALGEBRA. 181 after 4 years, $(\frac{21}{20})^4 a + (\frac{21}{20})^3 b + (\frac{21}{20})^2 b + \frac{21}{20} b + b$, after $n$ years, $(\frac{21}{20})^n a + (\frac{21}{20})^{n-1} b + (\frac{21}{20})^{n-2} b + \ldots + \frac{21}{20} b + b, \ldots$ This amount evidently consists of two parts, of which the first is $(\frac{21}{20})^n a$; and the other, taken inversely, forms the series $b + \frac{21}{20} b + (\frac{21}{20})^2 b + (\frac{21}{20})^3 b + \ldots + (\frac{21}{20})^{n-1} b$; which series is evidently a geometrical progression, the ratio of which is equal to $\frac{21}{20}$, and we shall therefore find its sum, by first multiplying the last term $(\frac{21}{20})^{n-1} b$ by the exponent $\frac{21}{20}$; which gives $(\frac{21}{20})^n b$. Then, subtracting the first term $b$, there remains $(\frac{21}{20})^n b - b$; and, lastly, dividing by the exponent minus 1, that is to say by $\frac{1}{20}$, we shall find the sum required to be $20(\frac{21}{20})^n b - 20b$; therefore the amount sought is, $(\frac{21}{20})^n a + 20(\frac{21}{20})^n b - 20b = (\frac{21}{20})^n \times (a+20b) - 20b$. 551. The resolution of this formula requires us to calculate, separately, its first term $(\frac{21}{20})^n \times (a+20b)$, which is $n \log. \frac{21}{20} + \log. (a+20b)$; for the number which answers to this logarithm in the Tables will be the first term; and if from this we subtract $20b$, we shall have the amount sought. 552. A person has a principal of 1000l. placed out at five per cent, compound interest, to which he adds annually 100l. beside the interest: what will be the amount of this principal at the end of twenty-five years? We have here $a=1000$; $b=100$; $n=25$; the operation is therefore as follows: $log. \frac{21}{20} = 0.021189299$ Multiplying by 25, we have $25 log. \frac{21}{20} = 0.5297324750$ $log. (a+20b) = 3.4771213135$ And the sum $= 4.0068537885$. So that the first part, or the number which answers to this logarithm, is 10159.1, and if we subtract $20b=2000$, we find that the principal in question, after twenty-five years, will amount to 8159.1l. 553. Since then this principal of 1000l. is always increasing, and after twenty-five years amounts to $8159\frac{1}{10}$l. we may require, in how many years it will amount to 1000000l. Let $n$ be the number of years required: and, since $a=1000$, $b=100$, the principal will be, at the end of $n$ years, $(\frac{21}{20})^n \cdot (3000) - 2000$, which sum must make 1000000; from it therefore results this equation; $3000 \cdot (\frac{21}{20})^n - 2000 = 1000000$;