a 1. The first term: 5. The sum of all the terms. Note. As the last term in a long series of numbers is very tedious to come at, by continual multiplication ; therefore, for readier finding it out, there is a series of numbers made use of in Arithmetical 'Propor. tion, called indices, beginning with an unit, whose common difference is one ; whatever number of indices you inake use of, set as many numbers (in such Geometrical Proportion as is given in the puestion) inder them. As 1, 2, 3, 4, 5, 6 Indices. 2, ., 8, 16, 32, 64. Numbers in geometrical proportion. But if the first term in geometrical proportion be different from the ratio. the indices must begin with a cypher As 0, 1, 2, 3, 4, 5; 6. Indices. 1, 2, 4, 8, 16, 32, 64. Numbers in geometrical proportion. When the indices begin with a cypher, the sum of the indices made choice of must always be one less than the number of terms, given in the question ; for 1 in the indices is over the second term, and 2 over the third, ge. Add any two of the indices together, and that sum will agree with the product of their respective terms, As in the first table of indices 2 + 5 = 4 X 16 = 64 In any geometrical progression proceeding from unity, then ratio being known, to tind any remote term, without producing all the intermediate terms. Rule. Find what figure of the indices added together would give the exponent of the term wanted; then multiply the numbers standing under such exponent into each other, and it will give the term required. Note. When the exponent 1 stands over the second term, the number of exponents must be 1 less than the number of terms. 1. A man agrees for 12 peaches, to pay only the price of the last, reckoning a farthing for the first, and a halfpenny for the second, &c. doubling the price to the last, what must he give for them? Ans. £%...2...8. 16= 4 0, 1, 2, 3, 4, Exponents. 16= 4 1, 2, 4, 8, 16. No of terms. 2568 8-3 6 EXAMPLES For 4+4+3=1. No. of terms less, 1 *)2048=11 Nu. of far. 12)512 210)4|2...8 £2...2...8 2. A country gentleman going to a fair to buy some oxen, meets with a person who had 23; he demanded the price of them, was answered £16 a piece: the gentleman bids him £15 a piece, and he would buy all; the other tells him it could not be taken ; but if he would give what the last ox would come to, at a farthing for the first and doubling it to the last, he should have all. What was the price of the oxen? Ans. £1369.......4 In any geometrical progression not proceeding from unity, the ratio being given, to find any remote term, without producing all the intermediate terms. Rule. Proceed as in the last, only observe that every product must be divided by the first term. 3. A sum of money is to be divided among 8 persons, the first to have £20, the second £60, and so in triple proportion ; what will the last have? Ans. £43740. 0, 1, 2, 3, 540X540 14510X 60. V), 60, 180, 540, 20 20 ---43740 3+3+1=7, one less than the number of terms. 4. A gentleman dying left nine sons, to whom and to his executore, he bequeathed his estate in manner following: To his executors £50. his youngest son was to have as much more as the executors, and each son to exceed the next younger by as much more; what was the eldest son's portion ? Ans. £85600. The first term, ratio, and number of terms given, to find the sum of all the terms. Rule. Find the last term as before, then subtract the first from it, and divide the remainder by the ratio, less 1 ; to the quotient of which add the greater, gives the sum required. 5. A servant skilled in numbers agreed with a gentleman tu serve him twelve months, provided he would give him a farthing for his first month's service, a penny for the second, 4d. for the third, &c. what did his wages amount to? Ans. £5825.8...57. 256 X 256=65536, then 65536 X 64=4194304. 0, 1, 2, 3, 4, 4194304-1 1, 4, 16, 64, 256. -1 -1398191, then ++4+3=11. No. of terms less 1, 1398101 +4194304=6592405 farthings. EXAMPLES 6. A man bought a horse, and by agreement was to give a farthing for the first nail, three for the second, &c. there were four shoes, and in each shoe 8 nails; what was the worth of the horse? Ans. £9651 14681693...13...4. 7. A certain person married his da:ighter on New-year's day, and gave her husband is. towards her portion, promising to double it on the first day of every month for 1 year; what was her portion ? Ans. £901...15. 8. A laceman, well versed in numbers, agreed with a gentleman to sell him 24 yards of rich gold brocąded lace, for % pins the first yard, 6 pins the second, &c. in treble proportion; i desire to know what he sold the lace for, if the pins were valued at 100 for a farthing; also what the laceman got or lost by the sale thereof, supposing the lace stood him in £7. per yard ? Ans. The lace sold for £326886...0...9. Guin £326732...0...9. EXAMPLES. PERMUTATION Rule. Multiply all the given terms one into another, and the last product will be the number of changes required. 1. How many changes may be rung upon 12 bells; and how long would they be ringing but once over, supposing 40 change, might be rung in 1 minute, and the year to contain 365 days, hours ? 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9 X 10 X11X12=479001600 changes, which-10=47900160 minutes ; and if reduced, is =91 years, 3 weeks, 5 days, 6 hours. 2. A young scholar coming into town for the convenience of a good library, demands of a gentleman with whom he lodged, what his diet would cosi for a year, who told him £10. but the scholar not being certain what time he should stay, asked him what he must give him for so long as he should place his family (consisting of 6 persons besides himself) in different positions, every day at dinner; the gentleman thinking it would riot be long, tells hiin £5, to which the scholar agreos. What time did the scholar stay with the gentleman ? Ans. 5040 days. (T. S.) In an establishment of 100 pupils, I desire to know how many different positions you can place them in at dinner, and what would be the amount of board and education for one young gentleman, provided he remained at school during the above changes, terms being 40 guineas per annum, and what would the master gain by his establishment, providel lie had the above number of pupils all the time of the changes; his profit being 10 per cent, and what would the pupils receive in pocket money, half of them to have 2d. I, 3d. ia, 4d. and the rest och per weck? THE TUTOR'S ASSISTANT. 9 PART II. VULGAR FRACTIONS. A FRACTION is a part or parts of an unit, and written with k, The figure above the line is called the numerator, and the under one the denominator; which shews how many parts the unit is divided into ; and the numerator shews how many of those parts are meant by the fraction. There are four sorts of vulgar fractions: proper, improper, compound, and mixed, viz. 1. A PROPER FRACTION is when the numerator is less than the denominator, as i, ż, ł, ile 170, fc. 2. AN IMPROPER FRACTION is when the numerator is equal to, or greater than the denominator, as i, i, , fc. 3. A COMPOUND FRACTION is the fraction of a fraction, and known by the word of, as d, of ý, of }, of 1,, of is, fc. 4. A MIXED NUMBER OR FRACTION is composed of a whole number and fraction, 8., 17}, 8 7. &c. REDUCTION OF VULGAR FRACTIONS. 1. To reduce fractisns to a common denominator, RULE. Multiply each numerator into all the denominators, except its own, for a new numerator; and all the denominators ' for a common denominator. Or, 2. Multiply the common denominator by the several given nurnerators separately, and divide their product by the several denominators, the quotient will be the new numerators. 12 197 EXAMPLES 16 1. Reduce and to a common denominator. Facit in, and ** 1st. num. 2d. num. 2X7 14 4X4 16, then 4 X7 28 den., and 16 2. Reduce $, 2, and f to a common denominator. Facit 7%, , 3. Reduce , 616, and $, to a common denominator. 2240 2016 7380 3367) 3360 1370) 1330 to Reduce 1 640 240 R40 Facst ICO) 1780 1280) 7680 Facit *240. 672 560 860 105 Facit 720 340 1299 5. Reduce , i, j, and } to a common denominator. Facit , 6. Reduce ž, i, j, and to a common denominator. 1200 116021703 37603 3137 2. To reduce a vulgar fraction to its lowest terms. Rule. Find a common measure by dividing the lower. term by the upper, and that divisor by the remainder following, till nothing remain ; the last divisor is the common measure; the divide both parts of the fraction by the common' measure, and the quotient will give the fraction required. Note. If the common measure happens to be onc, the fraction is als ready in its lowest term; and rohen a fraction hath cyphers at the right Pound, it may be abbreviated by eutting them off, as lo. EXAMPLES. Facit 52 171 7. Reduce si to its lowest terms. then (11)=Facit. Facit is. 9. Reduce 208 to its lowest terms. 10. Reduce 197 to its lowest terms. Facit . 11. Reduce 823 to its lowest terms. Facit 12. Reduce 5184 to its lowest terms. Facit 3. To reduoe a mixed number to an improper fraction. RULE. Multiply the whole number by the denominator of the fraction, and to the product add the numerator for a new numerator, which place over the denominator. Note. To express whole numbers fraction-ways, set 1 for the denominator given. EXAMPLES 13. Reduce 183 to an improper fraction. Facit im 18x7+3=129 new numerator,=120. 14. Reduce 5615 to an improper fraction. 15. Reduce 1831 to an improper fraction. 16. Reduce 13, to an improper fraction. 17. Reduce 27 to an improper fraction. 18. Reduce 514% to an improper fraction.. 4. To reduce an improper fraction to its proper terms. Rule. Divide the upper term by the lower. EXAMPLES. 19. Reduce to its proper terms. Facit 185 129-7=187. 20. Reduce 1245 to its proper terms. Facit 56700 21. Reduce 38 to its proper terms. Facit 1832 Facit 1948 Facit siis Facit 2. Fucit ? 245 Facit 8229 |