Elements of Algebra

184 ELEMENTS SECT. III. The natural number of this logarithm is found to be $100107$. So that, subtracting the principal, $100000$ from this amount, the interest, for eight days, is 107l. 560. To this subject belongs also the calculation of the present value of a sum of money, which is payable only after a term of years. For as 20l., in ready money, amounts to 21l. in a year; so, reciprocally, a sum of 21l., which cannot be received till the end of one year, is really worth only 20l. If, therefore, we express, by $a$, a sum whose payment is due at the end of a year, the present value of this sum is $\frac{20}{21}a$; and therefore to find the present worth of a principal $a$, payable a year hence, we must multiply it by $\frac{20}{21}$; to find its value two years before the time of payment, we multiply it by $(\frac{20}{21})^2 a$; and in general, its value, $n$ years before the time of payment, will be expressed by $(\frac{20}{21})^n a$. 561. Suppose, for example, a man has to receive for five successive years, an annual rent of 100l. and that he wishes to give it up for ready money, the interest being at $5$ per cent; it is required to find how much he is to receive. Here the calculations may be made in the following manner: For 100l. due after $1$ year, he receives $95.239$ after $2$ years - - - - $90.704$ after $3$ years - - - - $86.385$ after $4$ years - - - - $82.272$ after $5$ years - - - - $78.355$ Sum of the 5 terms = $432.955$ So that the possessor of the rent can claim, in ready money, only 432.955l. 562. If such a rent were to last a greater number of years, the calculation, in the manner we have performed it, would become very tedious; but in that case it may be facilitated as follows: Let the annual rent be $a$, which commencing at present, and lasting $n$ years, will be actually worth $a + (\frac{20}{21})a + (\frac{20}{21})^2 a + (\frac{20}{21})^3 a + (\frac{20}{21})^4 a + \ldots + (\frac{20}{21})^n a$. This is a geometrical progression, and the whole is reduced to finding its sum. We therefore multiply the last term by the exponent, the product of which is $(\frac{20}{21})^{n+1}a$; then, subtracting the first term, there remains $(\frac{20}{21})^{n+1}a - a$; and, lastly, dividing by the exponent minus 1, that is, by $-\frac{1}{21}$, or, which amounts to the same, multiplying by $-21$, we shall have the sum required, $-21 \cdot (\frac{20}{21})^{n+1} \cdot a + 21a$, or, $21a - 21 \cdot (\frac{20}{21})^{n+1} \cdot a$;