Any three of these being given, the other two may be found. The powers of the ratio are frequently denoted by figures placed over them, called indices : 1 2 3 4 Thus, 5, &c. Indices or Arithmetical series 2, 4, 8, 16, 32, &c. Geometrical series. CASE I. -The first term, ratio, and number of terms being given, to find the last term. RULE.-Raise the ratio to a power whose index shall be one less than the number of terms; multiply that power by the first term, and the product will be the last term or answer. Or, write down a few leading powers of the ratio, with their indices over them. Add the most convenient indices, to make an index one less than the number of the term sought. Multiply together the powers belonging to those indices, and their product multiplied by the first term, will be the answer, or term sought. EXAMPLES 1. A merchant bought 9 yards of broadcloth, and by agreement was to give what the last yard would come to, reckoning 4 cents for the first yard, 12 cents for the second, 36 cents for the third, and so on, to the last ; what did the cloth cost him, paying only the price of the ninth yard ? Ans. $262,44 cents. 3X3X3X3X3X3X3X3x4=26244. Ans. Or thus : DEM.-The reason of the rule is 1st power, 2d p. 3d p. 4th p. evident from the manner in which Ratio, 3, 9, 27, 81 a geometrical series is formed, be cause in the geometrical series, 2, 81 4, 8, 16, 32, &c. which has already 81 been given, the ratio being 2, it is 648 plain, that the second term is form ed by multiplying the first term by 8th power, 6561 ratio, the third term, by mul4 tiplying the second term by the ra tio, and so on; hence it is obvious, Ans. or 9th term, 26244 that the ratio raised to a power one less than the number of terms, and that power multiplied by the first term, must give the last term, because the same multipliers are used, and whether the first term be first or last used as a factor, can make no difference. 2. A sum of money was divided among 8 persons, the first 8 received $5, the second $15, the third $45, and so on, in hree fold ratio; the share of the eighth person is required ? Ans. $10,935, the share of the last 3. A farmer bargains for 16 sheep, to pay only the price of the last, reckoning one cent for the first, two cents for the second, four for the third, and so on, doubling the price to the last; how much must he pay for them? Ans. $327,68cts. CASE II.—The first term, the last term, and the ratio being given, to find the sum of all the terms or series. RULE.-Divide the difference of the extremes, that is, the difference. of the first and last terms, by the ratio, less 1, and the quotient, increase ed by the greater term, will give the answer or sum of all the terms. EXAMPLES. 1. If the first term be 5, the last term 3645, and the ratio 3; what is the sum of all the terms? Ans. 5465. 3645 the last term. DEM.-The reason of this rule is 5 the first term. obvious from the nature of a geome trical series; thus, if the ratio of any 2)3640 geometrical series be 2, the differ1820 ence of the greatest and least terms 3645 is equal to the sum of all the terms except the greatest ; let 2, Ans. 5465 sum of all the terms. 4, 8, 16, 32, be a geome trical series whose ratio is 2, and whose sum is 24+8+16+32=62, and the difference of the extremes 32–2=30, and it is plain that the difference of the extremes is equal to the sum of all the terms except the last, for 2+4+8+16=30; the last term added to the difference of the extremes makes 32-2= 30+32=62: Here dividing the difference of the extremes by the ratio, less one, could make no difference, because it would be dividing by 1; hence when the ratio is 3, the difference of the extremes is double the sum of all the terms except the greatest ; and when the ratio is 4, the difference of the extremes is triple the sum of all the terms except the last, and so on. 2. The daughter of a wealthy gentleman, marrying on a new year's day, he gave her a guinea, promising to triple it on the first day of each month in the year for her portion; what was the daughter's portion ? Ans. 265720 guineas. 3. What debt can be paid in a year, by paying one cent the first month, ten cents the second, and so on, increasing each month in a tenfold proportion? Ans. $1111111111,11 cents. NOTE.-First find the last terin as in Case I. then find the sum of the series. QUESTIONS ON GEOMETRICAL PROGRESSION. What is Geometrical Progression ? A. It is either increasing series of numbers by one common multiplier, or diminishing a number by one common divisor. When the first term, the ratio, and number of terms a åre given, how do you find the last term? A. By raising the ratio to a power one less than the number of terms, and multiplying that pow; er by the first term, the product will be the last term. When the first term, the last term, and the ratio are given, how do you find the sum of all the terms ? A. By dividing the difference of the extremes by the ratio, less one, and that quotient increased by the greater term, will be the sum of all the terms. POSITION, Is a rule for performing certain questions which cannot be worked by the common direct rules, until certain operations are performed, which are required to be done in the conditions of the questions.This rule is sometimes called False Position, or False Supposition, because there is a supposition of some false numbers made, to work with, the same as if they were the true ones, and by means of these suppositions, the true numbers are discovered. It is sometimes called Trial and Errour, because we proceed by trial of false numbers, and by them find out true ones by a comparison of the errours. Position is divided into two parts, Single and Double. SINGLE POSITION, Teaches to resolve such questions as require only one supposition. The result in this rule is always proportional to the supposition, RULE.—Take or assume any number for that which is required, and perform the same operations with it, as are described or performed in the question. Then say, as the result of the operation, is to the position, or number assumed, so is the result in the question, to a fourth term, which will be the answer to the question, or number sought. Proved, by adding the several parts of the sum together, and if it agrees with the given sum, the work is right. EXAMPLES. 1. What number is that, which being multiplied oy 7, and the product divided by 6, the quotient may be 42 ? Ans. 36. 84 the supposed number. DEM.--It is evident, that this 7 rule is nothing more than the Rule of Three, after the opera. 6)588 PROOF. tion is performed which is re98:84: : 42 36 quired to be performed in the conditions of the question, be42 7 cause it is obvious, that as the 168 6)252 result of the operation, 98, is to 336 the supposed number, 84, which 42 produced it, so is 42 to the true 98)3528(36 Ans. number which has produced 42 by the 294 same operation, which is evident from the proof. It is obvious that we here have a 588 plain statement in the Rule of Three, for 588 nc rem. as the result of the supposed number, is to the supposed number, so is the result wnich is given in the question, to the number that droduced it; hence the reason of our rule is evident. 2. A, B, and C, buy a quantity of cloth for $340; of which A pays three times more than B, and B four times more than C; what did each pay? Ans. A paid $240,—B, $80,—and C, $20. 3. A general, after sending out a foraging 1 and } of his men, had one thousand remaining; what number of men had he at first ? Ans. 6000. 4. A gentleman being asked his age, said, if } of the years I have lived, be multiplied by 7, and ß of them be added to the product, the sum will be 219; what was his age ? Ans. 45 years. 5. A cistern containing 600 gallons of cider, has three unequal pipes; the greatest pipe when opened, will empty the cistern in one hour, the second will empty it in two hours, and the third will empty it, in three hours; in what time will the three empty it all running at the same time? Ans. 32-4 min. 6. What sum of money, at 6 per cent per annum, will amount to $860 in ten years? Ans. $537,50 cents. 7. A gentleman bought a horse, chaise, and harness for $540, the harness cost half as much as the horse, and the horse and harness half as much as the chaise; what did each cost? Ans. Harness $60, horse $120, and chaise $360, 8. A gentleman being asked how much money he had on hand, said that }, }, , and of his money amounted to $114; what amount of money had he? Ans. $120. QUESTIONS ON SINGLE POSITION. What is Single Position ? A. It teaches to resolve such questions as require only one supposition.. How do you proceed in the work ? A. Assume any number, and with it perform the operations which are required to be performed in the conditions of the question; then say, as the result of the operation, is to the assumed number, so is the result in the question to the answer. Wherein does this rule differ from the Single Rule of Three? A. In nothing except the operation to be performed according to the conditions of the question, previous to the statement. DOUBLE POSITION, Is a rule for working certain questions by means of two suppositions of false numbers. The questions belonging to this rule, have results which are not proportional to their suppositions or numbers assumed; such are those, in which the number sought is increased or diminished by some given number, which is no known part of the number required. RULE I.-1. Take or assume any two convenient numbers, and proceed with each of them separately according to the conditions of the question, as in Single Position.--2. Place the results or errours, against the positions or supposed numbers, thus, and Pos. Er. if the errour be too great, mark it with t; and if too 6 30 small, with 3d. Multiply them crosswise; that is, the first posi- X tion by the last errour, and last position by the first 8 20 errour.--4. If the errours be alike, that is, both too small, or both too great, divide the difference of the products by the difference of the errours, and the quotient will be the answer.-5. If the errours be unlike, that is, one too small, and the other too great, divide the sum of the products by the sum of the errours, and the quotient will be the answer. Or, instead of the preceding rule, it would perhaps be better to use Rule ÍI. see below: NOTE.- This Rule is founded on this supposition: That the first errour is to the second, as the difference between the true and first supposed number, is to the difference between the true and second supposed number; when that is not the case, the exact answer cannot be found by this Rule. RULE II.-As the difference of the errours, if alike, or their sum, if unlike, is to either errour, so is the difference of the positions, to a fourth number, which being added to, or subtracted from, the position producing the errour, will give the answer, or number sought. Note. The student will readily determine whether the position producing the errour, is to be added or subtracted from the fourth proportional number, by considering whether the position should be added or subtracted to produce the desired effect in the errour. 1. A person bought 30 quarts of liquor for 50s., part whiskey, at 6d. per quart, and brandy at 3s. per quart; how much did he buy of each? Ans. whiskey, 16qts., brandy, 14qts. 1st Position.-8 quarts whiskey at 6d. per quart is! 4s. Then 22 quarts of brandy at 3s. per quart is. 66s. 70 The given sum. First Errour, 20+ 2d Position.--.qts. whiskey at 6d. per qt. is 3s. Then, 24qts. of brandv at 3s. per qt. is 72s. 75s. First Position, 8qts. The given sum, 50 Second Position, 6qts. The ad errour, 25+ Dif. of Pos. 2 The 1st errour, 20 Dif. of errours, 5 By Rule 1. 8 PROOF 25+ Brandy, 14 at 3s.=42s. 50s 120 200 120 5)80 16 50 |