GEOMETRICAL PROGRESSION.* Any rank or series of numbers continually increasing by a common multiplier, or decreasing by a common divisor, is said to be in Geometrical Progression. S 1, 2, 4, 8, 16, &c. is an increasing geometrical series. Thus 16, 8, 4, 2, 1, &c. is a decreasing geometrical series. There are five terms in Geometrical Progression, any three of which being given, the other two may be easily found. The common multiplier or divisor, by which the series is increased or decreased, is called the ratio. N. B.-In any rank or series of numbers, which increase or decrease by a common ratio, the product of the two extremes is equal to the product of any two means, equally distant from the said extremes, as in the series, 2, 4, 8, 16; the product of 16×2 is equal to the product of 8x4, each product being 32. When the number of terms is odd, the middle one supplies the place of two terms, as in the series 27, 9, 3, the product of 27 x3 is equal to the product of 9×9, each product being 81. Also, the product of any two means divided by either of the extremes will produce the other extreme, &c. &c. When the first term, the ratio, and number of terms are given, to find the last term, and the sum of all the series. RULE. 1. Raise the ratio to the power whose index is one less than the number of terms given in the question; which, being multiplied by the first term of the series, will give the last term or greater extreme. 2. Multiply the last term by the ratio, and from the product subtract the first term of the series; then divide the remainder by the ratio, less one, and the quotient will be the sum of all the series. EXAMPLES. 1. If the first term of a geometrical series be 5, the ratio 3, and the number of terms 7, what is the last term, and the sum of the series. Geometrical Progression will admit of 18 problems. See Mr. Nicholas Pike's large Arithmetic, second edition, 1797. Multiply by 3=the ratio. 729 the 6th power of the ratio. 5the first term of the series. Answer 3645=the last term of the series. Multiply by 3 the ratio. 10935 Subtract 5 The 1st term of the series. The ratio=3-1=2)10930 Answer 5465=the sum of the series. 2. If the first term of a geometrical series be 4 and the ratio 4, what is the 9th term only? 4 The ratio=4, and the 8th power 3. A gentleman by his will left his estate to his five sons, in the following manner, that is, to his youngest son $1000, to his second $1500, and ordered that each son should exceed the next younger by the equal ratio of 1; what was the whole amount of the gentleman's estate? 4 4. A man bought a horse, and by agreement was to give a farthing for the first nail, two for the second, four for the third,&c.; there were 4 shoes and 8 nails in each shoe; I demand what the horse is worth at that rate? Ans. 4473924£ 5s. 3 d. 5. A thresher wrought 20 days and received for the first day's labor 4 grains of wheat; for the second 12; for the third 36, &c. How much did his wages amount to, allowing 7680 grains to make a pint, and the whole to be disposed of at $1 per bushel? Ans. $14187.75cts. 6. A gentleman, whose daughter was married on a new year's day, gave her a guinea, promis ing to triple it on the first day of each month in the year; what did her portion amount to? Ans. 265720 guineas. 7. An ignorant fop wanting to purchase an elegant house, a facetious gentleman told him he had one which he would sell upon these moderate terms, namely: that he should give him one cent for the first door, two for the second, four for the third, and so on, doubling at every door, which were 36 in all. It is a bargain, cried the simpleton, and here is a dollar to bind it. How much did the house cost him? Ans. $687194767.35cts. 8. A crafty servant agreed with a farmer (ignorant in numbers) to serve him 12 years, and to have nothing for his service but the produce of a wheat corn for the first year, and that produce to be Bowed for the second year, and so on from year to year till the end of the said time. I demand the worth of the whole produce, allowing the increase to be but in a tenfold proportion, and sold out at 50 cents per bushel? Ans. $113028. 9. A man threshed wheat 9 days for a farmer, and agreed to receive but 8 wheat corns for the first day's work, 64 for the second, and so on, in an eightfold proportion. What did his 9 days' labor amount to, rating the wheat at 833cts. per bushel? Ans. $260, rejecting remainders. 10. A merchant sold 30 yards of fine silk velvet, trimmed with gold very curiously, at 2 pins for the first yard, 6 pins for the second, 18 pins for the third, &c., in triple proportion. I demand how much the velvet produced when the pins were afterwards sold at 100 for a farthing; also, whether the merchant gained or lost by the transaction, and how much, supposing the velvet to have cost him 100 by the yard? Ans. The velvet produced 2144699292€ 13s. Old., and the merchant gained 2144696292€ 13s. Ožd. PERMUTATION. The permutation of quantities is the showing how many different ways any given number of things may be changed, so as to assume different positions: thus-abe, acb, bac, bca, cab, cba, are six diffe rent positions of three letters. CASE 1. To find the number of permutations or changes that may be made of any number of things, all different from each other. RULE. Multiply all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product will be the number of permutations required. EXAMPLES. 1. How many changes of position can a company of 6 men assume? 1×2×3×4×5×6=720 Ans. 2. How many different numbers can be made with 123456789 ? 1×2×3×4×5×6×7×8×9=362880. Ans. 3. How many changes may be rung upon 12 bells, and how long will they be ringing but once over, supposing 24 changes to be rung in one minute, and the year to contain 365 days and 6 hours? 1x2×3×4×5×6×7×8×9×10×11=39916800 12 24)479001600=changes. 60) 19958400-minutes. 365 days 6 hours=8766 hours)332640hrs.(37 years 24)8298 hours rem. 7)345 days 18 hours. 49 weeks 2 days; consequently, the number of changes is 479001600, and the time is 37 years, 49 weeks, 2 days, and 18 hours. 4. Seven gentlemen that were travelling met together by chance at a certain inn, upon the road, where they were so well pleased with their host and each other's company, that, in a frolic, they offered him $100 to let them stay at that place so long as they, together with him, could sit every day at dinner in a different order. The host, thinking that they could not sit in many different positions, because there were but few of them, and that himself could make no considerable alteration, he being but one, imagined that he should make a good bargain, and readily (for the sake of a good dinner and better company) entered into an agreement with them, and so made himself the eighth person. I demand how many different positions they sat in, and how long they staid at the said inn, allowing the year to be 365 days 6 hours? Ans. They sat in 40320 positions, and staid at the said inn 110 years, 142 days, and 12 hours. CASE 2. Any number of different things being given, to find how many changes can be made out of them, by taking any given number of quantities at a time. RULE. Take a series of numbers, beginning at the number of things given, and decreasing by 1, to the number of quantities to be taken at a time, and the product of all the terms, taken as above directed, will be the answer required. Any number of things being given, whereof there are several things of one sort, several of another sort, &c., to find how many changes may be made out of them all. RULE. 1. Take the series 1, 2, 3, 4, &c. up to the whole number of things given, and find the product of all the terms. 2. Take the series 1, 2, 3, 4, &c. up to the number of things given of the first sort, and do the same by the second, third, &c. sorts. 3. Divide the product of all the terms by the joint product of the different terms, and the quotient will be the answer. EXAMPLES. 1. How many changes or variations can be made out of the letters in the word Zaphnathpaaneah? There are fifteen letters in the given word; therefore, the number of things given is 15-consequently 1x2x3x4x5×6×7×8×9 × 10 × 11 × 12 x 13× 14 × 15= 1307674368000=the product of the whole number of terms-and 1×2×3×4×5 (=the number of ays = 120 = 2. 6. 2 1x2x3 (the number of aitches)= Now, 120×2×6×2=2880=the divisor. 2880) 1307674368000(454053600 Ans. 2. How many different numbers can be made of the following figures 1223334444 ? Ans. 12600. |