(2.) Three persons are to pay a reckoning of 20s.; A is to pay, B, and C; how much must each person pay of the reckoning?

(3.) A can do a piece of work in 7 days, B can do the same in 5, and C in 6. Set them all at work together, in what time will they finish it? .

I

(4.) One-fifth part of an army were killed in battle, part were taken prisoners, and part died by sickness; if 4000 men were left, how many men did the army at first consist of?

(5.) I have a cistern which has three cocks, D, E, and F. Now, if D be opened by itself, when the cistern is full, it will empty it in 9 hours; if E be opened by itself, it will empty the cistern in 11 hours; and, if F be opened by itself, it will empty the cistern in 13 hours. In what time will they empty the cistern if I set them all open together?

(6.) A person delivered to another a sum of money, to receive interest for the same at 4 per cent. per annum, (simple interest.) At the end of 3 years he received for principal and interest 1761. 8s. What was the sum lent?

DOUBLE POSITION.

Definition. By Double Position, or two suppositions, are solved those questions wherein the errors are proportional to the difference between the true number, and each supposition.

RULE.

Suppose any two convenient numbers, and proceed with them according to the nature of the question, marking the errors (with + or -) according as they exceed or fall short of the truth.

Then,

Multiply the first supposition by the second error, and the second supposition by the first error, and divide the sum of the products by the sum of the errors, if they are differently marked; or the difference of the products by the difference of the errors, if they are marked alike, and the quotient will be the number sought.

Or, II.

Multiply the difference between the two supposed numbers by the less error, and divide the product by the sum of the errors, if they are differently marked; or by the difference if they are marked alike; and the quotient will be a correction of the number belonging to the less error, and must be added to it, if that error be less than the truth, or subtracted, if it be greater.

Note 1. Mr. Ward, Mr. Malcolm, and several other writers, have omitted the rule of Position, because all questions that can be solved by it are more readily solved by a simple equation in algebra. Though this observation be true, yet Position has its use; for it may frequently be applied to the solution of adfected and exponential equations in algebra, better than any other method, (particularly the second rule,) for, by repeating the process, the answer will continually approximate to the true number within any assigned degree of exactness. For this reason it is of essential service in the more abstruse parts of the-mathematics; for, in many difficult problems, there is hardly any other way to obtain a solution.

2. In any enquiry, where it is possible to prove the truth of the answer, when discovered. Make two suppositions as near the truth as you are able to guess, and from them deduce an answer, as directed in the second rule; then take the nearer of the two suppositions; and, the result above gained, as two other suppositions; and, in like manner, deduce another answer: proceed thus, till you have obtained an answer sufficiently exact.

Examples.

(1.) What number is that, which, being multiplied by 3, the product increased by 4, and that sum divided by 8, the quotient may be 32?

108-12X9 27+9

108

27+9=36)3024(84 answer.

144

By Rule 2.

=24, correction of the number (108) belonging to the

less error. Hence 108-24-84, as before.

(2.) A man has 2 excellent horses; and a single-horse chaise and furniture worth 150%. Now, if the first horse be put in the chaise, his value with the furniture, &c. will be three times that of the second horse without it; but, if the second horse be put in the chaise, his value will be double that of the first. What are the horses worth?

(3.) A person being asked the time of the day, replied, the day is now 16 hours long, and the sun rises at 4 o'clock. Now, if you add of the hours that have passed since the sun rose to 4 of those which must elapse before the sun sets, you will have the exact time of the day.

(4.) A person received 11 crowns and 7 dollars for a debt of 47. 10s. 10d., and at another time received 4 crowns and 3 dollars for a debt of 11. 15s. What was the value of a crown and of a dollar in English money?

(5.) A person distributed in charity 2d. a piece among several poor children, and had 4d. left. He would have given them 3d. a piece, but wanted 10d. to be able to do it. What was the number of children?

Examples exercising Note 2.

(6.) If g×4372—g)=4x4732-4 what is g?

Answer, 2916.

(7.) If r3+12r=300, what is r? Answer, 6'098, &c. (8.) Given a 100 to find the value of x.

Note. By the 2d note, and a table of logarithms the most intricate adfected equations in algebra may be solved.

ARITHMETICAL PROGRESSION.

Definition. When a series of numbers increase, or decrease, by an equal excess or difference, those numbers are said to be in arithmetical progression; such as, 2, 4, 6, 8, 10, &c.; or 15, 14, 13, 12, 11, &c. and the numbers which form such series are called the terms of the progression. The first and last terms are usually called the extremes.

Note 1. If three numbers be in arithmetical progression, the sum of the extremes will be equal to double the mean; and the product of the extremes, increased by the square of the common difference, will be equal to the square of the mean. Thus, if 5. 7. 9. be in arithmetical progression, then will 5+9=7×2, and 9×5+2×2—7×7.

2. If four numbers be in arithmetical progression, the sum of the two extremes will be equal to the sum of the means.

Thus, if 2. 5. 8. 11. be in arithmetical progression,

Then will 2+11=5+8.

3. If a series of numbers, (consisting of any number of terms) be in arithmetical progression, the sum of the extremes will always be equal to the sum of any two means equidistant from the extremes; or to double the mean, if the terms be odd.

Thus, if 3. 5. 7. 9. 11. 13. &c. be in arithmetical progression,
Then will 3+15=5+11=7+9.

Or, if 1. 4. 7. 10. 13. &c. be in arithmetical progression, then will 1+13-4+10=7×2.

4. The difference between the extremes is equal to the product of the common difference by the number of terms less one,

Thus, if 3. 5. 7. 9. &c. be in arithmetical progression,

Then will 9-3-2×4-1.

5. The number of terms, where the terms are odd, is equal to the sum of the terms, divided by the mean.

Thus, 3+5+7+9+11=35÷7=5.

6. The sum of the terms is equal to the number of terms multiplied by the mean term.

Thus 3+5+7+9+11=5×7.

7. If out of any series of numbers in arithmetical progression there be taken any series of equidistant terms, that series will also be in arithmetical progression.

Thus, if 2. 4. 6. 8. 10. 12. 14, &c. be in arithmetical progression, Then will 4. 8. 12. &c. be in arith. prog.

8. In any series of numbers in arithmetical progression, the common excess or difference is as often repeated as there are terms in the progression, wanting one; viz. every term except the first is continually increased or diminished by the common excess or difference.

9. The rules for an ascending or descending series are the same; for, a descending series becomes an ascending one by beginning at the least term.

Proposition 1. Given the least term, the greatest term, and the number of terms, of an arithmetical progression, to find the sum of the terms.

Rule. To the least term add the greatest, multiply the sum by half the number, of terms, and the product will be the sum of the terms.

Prop. 2. Given the least term, the greatest term, and the number of terms, to find the common excess, or difference.

Rule. Divide the difference between the greatest and the least term by the number of terms less unity, and the quotient will be the common excess, or difference.

Prop. 3. Given the least term, the greatest term, and the common excess, or difference, to find the number of

terms.

Rule. Divide the difference between the greatest and the least term, by the common excess, or difference, the quotient, increased by an unit, will give the number of

terms.

Prop. 4. Given the greatest term, the number of terms, and the common excess, or difference, to find the

least term.

Rule. Multiply the common excess, or difference, by the number of terms less 1: subtract the product from the greatest term, and the remainder will be the least

term.

Prop. 5. Given the number of terms, the common excess, or difference, and the sum of the terms, to find the

least term.

Rule. Divide the sum of the terms by the number of terms; and, from the quotient, subtract half the product of the common excess, or difference, by the number of terms less 1, the remainder will be the least term.