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2. Suppose 4ool. are to be paid at the end of 2 years, and 2100l. at the end of 8 years ; what is the equated time for one payment, reckoning 5 per cent. simple inter
Ans. 7 years.
3. Suppose 300l. are to be paid at one year's end, and 300l. more at the end of 1 year ; it is required to find the time to pay it at one payment, 5 per cent. simple interest being allowed.
Ans. I'248637 year.
COMPOUND INTEREST is that, which arises from the principal and interest taken together, as it becomes due, at the end of each stated time of payment.
1. Find the amount of the given principal, for the time of the first payment by simple interest.
The equated time for any number of payments may be readily found when the question is proposed in numbers, but it would not be easy to give algebraic theorems for those cases, on account of the variation of the debts and times, and the difficulty of finding between which of the payments the equated time would happen. Supposing r to be the amount of il. for one year, and the oth
log. ar 46 er letters as before, then t
will be a general theorem
log. r. for the equated time of any two payments, reckoning compound interest, and is found in the same manner as the former.
* The reason of this rule is evident from the definition, and the principles of simple interest.
2. Consider this amount as the principal for the second payment, whose amount calculate as before, and so on through all the payments to the last, still accounting the last amount as the principal for the next payment.
1. What is the amount of 3201. 1os. for 4 years, at 5 per cent. per annum, compound interest ? 2.)3201. 105. Ist year's principal. 16 6d.
ist year's interest.
2. . What is the compound interest of 7601. 1os. forborn 4 years at 4 per cent. ?
Ans. 1291. 35. 6 d. 3. What is the compound interest of 410l. forborn for 25 years, at 4 per cent. per annum ; the interest payable half-yearly ?
Ans. 481. 45. 11 d. 4. Find the several amounts of 50l. payable yearly, half-yearly and quarterly, being forborn 5 years, at 5 per cent. per annum, compound interest.
Ans. 631. 16s. 3 d. 641. and 641. 1$. 9, d.
Letr = amount of il. for one year, and p = principal or given sum ; then since r is the amount of il, for one year, qoz will be its amount for two years, p} for 3 years, and so on ; for, when the rate and time are the same, all principal sums are necessarily as their amounts; and consequently as r is the principal for the second year, it will be as 1 :r:: 7 : ;? = amount for the second year, or principal for the third ; and again, as 1 :r::p? : ri= amount for the third year, or prin. cipal for the fourth, and so on to any number of years. And if the number of years be denoted by t, the amount of il. fort years will be ml. Hence it will appear, that the amount of any other principal sum p for 6 years is pri ; for as I : :; : pr', the same as in the rule.
If the rate of interest be determined to any other time than a year, as • , &c. the rule is the same, and then t will represent that stated time.
r= amount of 1l. for one year, at the given rate
t = time.
( m = amount for the time t. Then the following theorems will exhibit the solutions of all che cases in compound interest. I. pr=m.
20 Juvolve the amount thus found to such a power, as is denoted by the number of years.
3. Multiply this power by the principal, or given sum, and the product will be the amount required.
The most convenient way of giving the theorem for the time, as well as for all the other cases, will be by logarithms, as follows: I. tx log. r +log. p=log. m. II. log. m-t x log. r=log. p. log. m-log.
log. m-log. P III.
=log. r. log.r If the compound interest, or amount of any sum, be required for the parts of a year, it may be determined as follows :
I. When the time is any aliquot part of a year.
1. Find the amount of il. for one year, as before, and that root of it, which is denoted by the aliquot part, will be the amount sought.
2. Multiply the amount thus found by the principal, and it will be the amount of the given sum required.
II. When the time is not an aliquot part of a jear.
RULE. 1. Reduce the time into days, and the 365th root of the amount of 1l. for one year is the amount for one day.
2. Raise this amount to that power, whose index is equal to the number of days, and it will be the amount of 11. for the given time.
3. Multiply this amount by the principal, and it will be the amount of the given sum required.
To avoid extracting very high roots, the same may be done by logarithms thus : divide the logarithm of the rate, or amount of il. for one year, by the denominator of the giren aliquot part, and the quotient will be the logarithm of the root sought.
4. Subtract the principal from the amount, and the remainder will be the interest.
1. What is the compound interest of gool. for 4 years, at 5 per cent. per annum ?
1:05 = amount of il. for one year at 5 1'05
55125 22050 110250 11025
1'21550625=4th power of 1'05.
607*75312500 = amount.
107*753125 = 1071. 155. 02 d. = interest required.
2. What is the amount of 7601. 1os. for 4 years, at 4
Ans. 8891. 135. 6 d.
per cent. ?
3. What is the amount of 7211. for 21 years, at 4 per cent. per annum ?
Ans. 16421. 198. rod.
4. What is the amount of 2171. forborn 25 years, at 5 per cent. per annum, supposing the interest payable quar
Ans. 2421. 135. 4 d.