3. How many strokes do the clocks of Venice strike in the compass of the day, which go right on from 1 to 24 o'clock ? Ans. 300. 4. What debt can be discharged in a year, by weekly payments in arithmetical progression, the first payment being Is., and the last or 52d payment 57. 3s. ? Ans. 1351. 4s. PROB. IL Given the extremes, and the number of terms; to find the common difference. RULE-Subtract the less extreme from the greater, and divide the remainder by 1 less than the number of terms, for the common difference. EXAMPLES. 1. The extremes being 3 and 19, and the number of terins 9; required the common difference? 2. If the extremes be 10 and 70, and the number of terms 21; what is the common difference, and the sum of the series ? 16 Or, 19-3=18=2. Ans. the com. diff. is 3, and the sum is 840. 3. A certain debt can be discharged in one year, by weekly payments in arithmetical progression, the first payment being 1s., and the last 57. 3s.; what is the common difference of the terms? Ans. 2. PROB. III. Given one of the extremes, the common difference, and the number of terms; to find the other extreme, and the sum of the series. RULE.-Multiply the common difference by 1 less than the number of terms, and the product will be the difference of the extremes: therefore add the product to the less extreme, to give the greater; or subtract it from the greater to give the less. 2 8 EXAMPLES. 1. Given the least term 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series? 16 3 19 the greatest term. 3 the least. 22 sum. 9 number of terms. 2) 198 99 the sum of the series. 2. If the greatest term be 70, the common difference 3, and the number of terms 21; what is the least term and the sum of the series? Ans. the least term is 10, and the sum is 840. 3. A debt can be discharged in a year, by paying Is. the first week, 3s. the second, and so on, always 2s. more every week; what is the debt, and what will the last payment be? Ans. the last payment will be 57. 3s., and the debt is 135l. 4s PROB. IV. To find an arithmetical mean proportional between two given terms. RULE. Add the two given extremes or terms together, and take half their sum for the arithmetical mean required. Or, subtract the less extreme from the greater, and half the remainder will be the common difference; which being added to the less extreme, or subtracted from the greater, will give the mean required. EXAMPLE. To find an arithmetical mean between the two numbers 4 and 14. Or, 14 Or, 14 4 2) 18 2) 10 Ans. 9 EXAMPLE. To find two arithmetical means between 2 and 8. Here 8 com. dif. 2 5 the com. dif. 9 So that 9 is the mean required by both methods. PROB. V. To find two arithmetical means between two given extremes. RULE.-Subtract the less extreme from the greater, and divide the difference by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, gives the means. Then 2 + 2 EXAMPLE. To find five arithmetical means between 2 and 14. Here 14 lolo 5 4 the one mean, PROB. VI. RULE. To find any number of arithmetical means between two given terms or extremes. Subtract the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required. Then by adding this com. dif. continually, 6) 12 com, dif. 2 Note. More of Arithmetical Progression is given in the Algebra. GEOMETRICAL PROPORTION AND PROGRESSION. THE most useful part of Geometrical Proportion, is contained in the following theorems: THEOREM I.—If four quantities be in geometrical proportion, the product of the two extremes will be equal to the product of the two means. Thus, in the four 2, 4, 3, 6, it is 2 × 6 = 3 × 4 = 12. And hence, if the product of the two means be divided by one of the extremes, the quotient will give the other extreme. So, of the above numbers, the product of the means 12 ÷ 2 = 6 the one extreme, and 12÷ 6 2 the other extreme; and this is the foundation and reason of the practice in the Rule of Three. = THEOREM 2.—In any continued geometrical progression, the product of the two extremes is equal to the product of any two means that are equally distant from them, or equal to the square of the middle term when there is an uneven number of terms. Thus, in the terms 2, 4, 8, it is 2 × 8 = 4 × 4 = 16. it is 2 x 123 = 4 × 64 = 8 × 32 = 16 × 16 = 256. THEOREM 3.—The quotient of the extreme terms of a geometrical progression, is equal to the common ratio of the series raised to the power denoted by I less than the number of the terms. So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common ratio is 2, one less than the number of terms 9; then the quotient of the ex1024 tremes is 512, and 2o 512 also. 2 Consequently the greatest term is equal to the least term multiplied by the said power of the ratio whose index is 1 less than the number of terms. THEOREM 4.-The sum of all the terms, of any geometrical progression, is found by adding the greatest term to the difference of the extremes divided by 1 less than the ratio. So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, (whose ratio is 2,) is 1024 1 2 1024 + 2 1 1024+ 1022 = 2046. The foregoing, and several other properties of geometrical proportion, are demonstrated more at large in the Algebraic part of this work. A few examples may here be added of the theorems, just delivered, with some problems concerning mean proportionals. EXAMPLES. 1. The least of ten terms, in geometrical progression, being 1, and the ratio 2; what is the greatest term, and the sum of all the terms? Ans. the greatest term is 512, and the sum 1023. 2. What debt may be discharged in a year, or 12 months, by paying 1. the first month, 24. the second, 47, the third, and so on, each succeeding payment being double the last; and what will the last payment be? Ans. the debt 40957., and the last payment 20487. PROB. I. To find one geometrical mean proportional between any two numbers. RULE.-Multiply the two numbers together, and extract the square root of the product, which will give the mean proportional sought. Or, divide the greater term by the less, and extract the square root of the quotient, which will give the common ratio of the three terms: then multiply the less term by the ratio, or divide the greater term by it, either of these will give the middle term required. EXAMPLE. To find a geometrical mean between the two numbers 3 and 12. First way. 3 ) 12 ( 4, its root is 2 the ratio. 12 3 36 ( 6 the mean. 36 Then, 3 X 2 = 6 the mean. PROB. IL To find two geometrical mean proportionals between any two numbers. RULE.-Divide the greater number by the less, and extract the cube root of the quotient, which will give the common ratio of the terms. Then multiply the least given term by the ratio for the first mean, and this mean again by the ratio for the second mean: or, divide the greater of the two given terms by the ratio for the greater mean, and divide this again by the ratio for the less mean. EXAMPLE. To find two geometrical mean proportionals between 3 and 24. Here, 3) 24 ( 8, its cube root, 2 is the ratio. 12, the two means. PROB. III. To find any number of geometrical mean proportionals between two numbers. RULE. Divide the greater number by the less, and extract such root of the quotient whose index is 1 more than the number of means required, that is, the 2d root for 1 mean, the 3d root for 2 means, the 4th root for 3 means, and so on; and that root will be the common ratio of all the terms. Then with the ratio multiply continually from the first term, or divide continually from the last or greatest term. EXAMPLE. To find four geometrical mean proportionals between 3 and 96. Here, 3) 96 (32, the 5th root of which is 2, the ratio. Then, 3 x 2 = 6, and 6 × 2 = 12, and 12 × 2 = 24, and 24 × 2=48. Or, 96 2 48, and 48 2: 24, and 242≈ 12, and 12 ÷ 2 = 6. That is, 6, 12, 24, 48, are the four means between 3 and 96. OF MUSICAL PROPORTION. THERE is also a third kind of proportion, called Musical, which being but of little or no common use, a very short account of it may here suffice. Musical Proportion is when, of three numbers, the first has the same proportion to the third, as the difference between the first and second, has to the difference between the second and third. As in these, 6, 8, 12, 18; 18 :: 8 6: 18 - 8, When four numbers are in Musical Proportion; then the first has the same Proportion to the fourth, as the difference between the first and second has to the difference between the third and fourth. When numbers are in Musical Progression, their reciprocals are in Arithmetical Progression; and the converse, that is, when numbers are in Arithmetical Progression, their reciprocals are in Musical Progression. So, in these Musicals 6, 8, 12, their reciprocals, §, 1, are in arithmetical progression; for } + {= {;= 4; and j +}=}=; that is, the sum of the extremes is equal to double the mean, which is the property of arithmeticals. The method of finding out numbers in Musical Proportion, is best expressed by letters in Algebra. FELLOWSHIP OR PARTNERSHIP. FELLOWSHIP is a rule, by which any sum or quantity may be divided into any number of parts, which shall be in any given proportion to one another. By this rule are adjusted the gains, or losses, or charges of partners in company; or the effects of bankrupts, or legacies in case of a deficiency of assets or effects; or the shares of prizes, or the numbers of men to form certain detachments; or the division of waste lands among a number of proprietors. Fellowship is either Single or Double. It is Single, when the shares or portions are to be proportional each to one single given number only; as when the stocks of partners are all employed for the same time and Double, when each portion is to be proportional to two or more numbers; as when the stocks of partners are employed for different times. |