is the premium or money paid for the loan or use of money; and is distinguished into two kinds, simple and compound:

Simple interest is that which is paid for the principal, or sum lent, at a certain rate or allowance made by law, or agreement of parties, whereby so much as 5l. or 6l. or any other sum, is paid for 100l. lent out for one year; and more or less proportionally for greater or lesser sums, and for more or less time. For example, if it is 5l. to 100l. for one year, it is 2l. 10s. for half a year, and 10l. for two years: also 10l. for one year of 200l. and 5l. for half a year; and so on, for other sums and times. Thus, as the law, or agreement of parties, fixes a certain ratio, or, as we call it, rate of interest, which is so much on the 100l. for one year; from this we can easily find the proportional interest on 1l. for one year, being plainly the \(\frac{1}{100}\) part of the interest of 100l. so if this is 5l. that is .05l. if this is 6l. that is .06l. and if this is 5l. 10s. or 5.5l. that is .055l. Therefore, if we understand the rate of interest to be the interest of 1l. for one year, the more common questions about simple interest will relate to these four things, viz. any principal sum, its interest, the time in which it gives that interest, and the rate, or interest of 1l. for one year; according to which, that principal, interest, and time, are adjusted to one another.

From which we have four problems; in the rules whereof we suppose the principal and interest expressed in the denomination of pounds, by reducing what is less than 1l. to a decimal of 1l. and the time to be expressed in years, and decimal parts of one year.

Prob. I. Having any principal, sum, and time, with the rates of interest given, to find the interest of that sum for that time and rate.

Rule: Multiply the principal rate, and time, continually into one another; the product is the interest sought.

Observe, if we express the principal by \(p\), the interest by \(n\), the time by \(t\), and the rate by \(r\), then this rule is thus represented, \(n = prt\).

Example: The rate of interest being .05l. what is the interest of 85l. for 4 years and 3 quarters, or 4.75 years?

Answer: 20l. 3s. 9d. = 20.1875l. = 85 \times 4.75 \times 0.05.

Which is thus performed:

\[ \begin{align*} 85 &= p \\ 4.75 &= t \\ \end{align*} \]

\[ \begin{align*} 425 \\ 595 \\ 340 \\ \end{align*} \]

\[ \begin{align*} 403.75 \\ .05 &= r \\ \end{align*} \]

20.1875 pounds.

Which decimal is reduced by multiplying it by 20, 12, and 4: thus,

\[ \begin{align*} 1.875 \\ 20 \\ \end{align*} \]

\[ \begin{align*} 3.7500 \text{ shillings} \\ 12 \\ \end{align*} \]

\[ \begin{align*} 15000 \\ 7500 \\ \end{align*} \]

9,0000 pence

Prob. II. Having the rate, principal and interest, to find the time.

Rule: Divide the interest by the product of the rate and principal, the quote is the time: thus, \(t = \frac{n}{rp}\).

Example: The rate .05l. principal 85l. interest 20l. 3s. 9d. or 20.1875l. the time is 4.75 years, or 4 \(\frac{3}{4}\) years. Thus, 4.75 = \(\frac{20.1875}{85 \times 0.05}\), or \(\frac{20.1875}{4.25}\).

Demonstration: This rule is deduced from the former; thus, since \(n = trp\), then dividing both sides by \(rp\), it is \(\frac{n}{rp} = t\).

Prob. III. Having the principal, interest, and time, to find the rate.

Rule: Divide the interest by the product of principal and time, the quote is the rate: thus, \(r = \frac{n}{tp}\).

Example: \(n = 20.1875l.\), \(t = 4.75\) years, \(p = 85l.\), then is \(r = \frac{20.1875}{4.75 \times 85}\), or \(\frac{20.1875}{403.75}\).

Demonstration: Since \(n = trp\), divide both by \(tp\); it is \(\frac{n}{tp} = r\).

Prob. IV. Having the rate, time and interest, to find the principal.

Rule: Divide the interest by the product of rate and time, the quote is the principal; thus, \(p = \frac{n}{tr}\).

Example: \(n = 20.1875l.\), \(t = 4.75\) years, \(r = .05l.\), then is \(p = \frac{20.1875}{4.75 \times 0.05}\), or \(\frac{20.1875}{2375}\).

Demonstration: Since \(n = trp\), divide both sides by \(tr\), the quote is \(\frac{n}{tr} = p\).

Scholium: If the interest of any sum for any time is added to the principal, this total or sum is called the amount, (viz. of the principal and its interest for that time.) And then from these four things, viz. the amount, which we call \(a\), the principal, the time, and and rate, arise four problems; for having any three of these, the fourth may always be found. Thus,

Prob. V. Having the principal, time, and rate, to find the amount.

Rule: Find the interest by prob. I., add it to the principal, the sum is the amount.

Thus, by prob. I., the interest is \(ptr\); therefore the amount is \(a = ptr + p\). The reason is evident.

Note: Because \(ptr = rt \times p\), and \(p = 1 \times p\); therefore \(rt + p = rt + 1 \times p = a\). And so the rule may be expressed thus: To the product of the rate and time add unity, and multiply the sum by the principal, the product is the amount.

Example: What is the amount of £246l. principal in 2 years and \(\frac{1}{2}\), or 2.5 years, the rate of interest being .05? Answer £246l. + 30.75l. = £276l. 15s. for the interest is \(= 246 \times 0.05 \times 2.5 = 30.75\). Or thus: \(0.05 \times 2.5 = 125l.\) to which add 1, it is \(1 + 125l.\), which multiplied by 276, produces £276.75l.

Prob. VI. Given the principal, amount, and time, to find the rate.

Rule: Take the difference betwixt the principal and amount, and divide it by the product of the time and principal, the quote is the rate; thus, \(r = \frac{a - p}{tp}\).

Example: Suppose \(a = 276.75l.\), \(p = 246l.\), \(t = 2.5\) years; then \(r = 0.05 = \frac{276.75 - 246}{2.5 \times 246} = \frac{30.75}{615}\).

Demonstration: Since by prob. V., \(a = trp + p\), take \(p\) from both sides, it is \(a - p = trp\); then divide both by \(tp\), it is \(r = \frac{a - p}{tp}\).

Prob. VII. Given the amount, principal, and rate, to find the time.

Rule: Take the difference of the amount and principal, and divide it by the product of the principal and rate, the quote is the time; thus \(t = \frac{a - p}{rp}\).

Example: Suppose \(a = 276.75l.\), \(p = 246l.\), \(r = 0.05\); then \(t = 2.5\) years \(= \frac{276.75l. - 246}{246 \times 0.05} = \frac{30.75}{123}\).

Demonstration: In the last problem, \(a - p\) was equal to \(trp\); and dividing both by \(rp\), it is \(r = \frac{a - p}{rp}\).

Prob. VIII. Given the amount, rate, and time, to find the principal.

Rule: Add 1 to the product of the rate and time, and by that sum divide the amount, the quote is the principal; thus, \(p = \frac{a}{rt + 1}\).

Example: \(a = 276.75l.\), \(r = 0.05\), \(t = 2.5\) years; then \(p = 246 = \frac{276.75}{2.5 \times 0.05 + 1} = \frac{276.75}{1.125}\).

Demonstration: By prob. V., it is \(a = rt + 1 \times p\); therefore dividing both sides by \(rt + 1\), it is \(p = \frac{a}{rt + 1}\).

Compound Interest, is that which is paid for any principal sum, and the simple interest due upon it for any time, accumulated into one principal sum. Example: if 100l. is lent out for one year at 6l. and if at the end of that year the 6l. due of interest be added to the principal, and the sum 106l. be considered as a new principal bearing interest for the next year (or whatever less time it remains unpaid) this is called compound interest, because there is interest upon interest, which may go on by adding this second year's interest of 106l. to the principal 106l. and making the whole a principal for the next year.

Now, although it be not lawful to let out money at compound interest, yet in purchasing of annuities or pensions, &c., and taking leaves in reversion, it is very usual to allow compound interest to the purchaser for his ready money; and therefore, it is very necessary to understand it.

Let therefore, as before, \(p\) = the principal put to interest; \(t\) = the time of its continuance; \(a\) = the amount of the principal and interest; \(R\) = the amount of 1l. and its interest for one year, at any given rate, which may be thus found.

Viz. 100 : 106 :: 1 : 1.06 = the amount of 1l. at 6 per cent. Or 100 : 105 :: 1 : 1.05 = the amount of 1l. at 5 per cent. And so on, for any other assigned rate of interest.

Then if

\(R\) = amount of 1l. for 1 year, at any rate.

\(R^2\) = amount of 1l. for 2 years.

\(R^3\) = amount of 1l. for 3 years.

\(R^4\) = amount of 1l. for 4 years.

\(R^5\) = amount of 1l. for 5 years.

Here \(t = 5\). For 1 : \(R\) :: \(R\) : \(RR\) :: \(RR\) : \(RRR\) :: \(RRR\) : \(R^4\) : \(R^5\) : &c. in a geometrical progression continued; that is, as 1l. is to the amount of 1l. at 1 year's end : so is that amount : to the amount of 1l. at 2 years end, &c. Whence it is plain, that compound interest is grounded upon a series of terms, increasing in geometrical proportion continued; wherein \(t\) (viz. the number of years) does always assign the index of the last and highest term, viz. the power of \(R\), which is \(R^t\).

Again, as 1 : \(R^t\) :: \(p\) : \(pR^t\) = the amount of \(p\) for the time, that \(R^t\) = the amount of 1l. That is, as 1l. is to the amount of 1l. for any given time :: so is any proposed principal, or sum : to its amount for the same time.

From what has been said, we presume, the reason of the followings theorems will be very easily understood.

Theorem I. \(pR^t = a\), as above.

From hence the two following theorems are easily deduced.

Theorem II. \(\frac{a}{R^t} = p\).

Theorem III. \(\frac{a}{p} = R^t\).

By these three theorems, all questions about compound interest may be truly resolved by the pen only, viz. without tables: though not so readily as by the help of tables calculated on purpose.

Example Example I. What will £561. 10s. amount to in 7 years, at 5 per cent. per annum, compound interest?

Here is given \( p = 256 \), \( t = 7 \), and \( R = 1.05 \), which being involved until its index \( i \) (viz. 7) will become \( R^i = 1.40710 \). Then \( 1.40710 \times 256.5 = 360.92115 \)

\[ a = 360.92115 \]

which is the answer required.

Example II. What principal or sum of money must be put out to raise a stock of £360l. 18s. 5d. in seven years, at 5 per cent. per annum, compound interest?

Here is given \( a = 360.92115 \), \( R = 1.05 \) and \( t = 7 \) to find \( p \) by theorem II. Thus \( R^t = 1.40710 \times 360.92115 = 256.5 = p \).

That is, \( p = 256l. 10s. \) which is the sum or principal required.

Example III. In what time will £561. 10s. raise a stock of (or amount to) £360l. 18s. 5d. allowing 5 per cent. per annum, compound interest?

Here is given \( p = 256 \), \( a = 360.92115 \), \( R = 1.05 \).

To find \( t \) by theorem III. \( R^t = \frac{a}{p} = \frac{360.92115}{256} = 1.40710 \).

Which being continually divided by \( R = 1.05 \) until nothing remain, the number of those divisions will be \( t = 7 \).

Thus \( 1.05 \times 1.40710 \times 1.3400 \), and \( 1.05 \times 1.3400 \times 1.2762 \), and \( 1.05 \times 1.2762 \times 1.2155 \), and so on until it becomes \( 1.05 \times 1.05 \times 1 \), which will be at the seventh division.

Therefore it will be \( t = 7 \), the number of years required by the question.