2. Geometrical Progression. 280. A Geometrical Progression is a series of terms which increase by a constant multiplier, or decrease by a constant divisor, as 2, 4, 8, 16, 32, &c., increasing by the constant multiplier, 2, or 27, 9, 3, 1, }&c., decreasing by the constant divisor, 3. The multiplier or divisor, by which the series is produced, is called the ratio. 281. A person bought 6 brooms, giving 3 cents for the first, 6 cents for the second, 12 for the ihird, and so on, doubling the price to the sixth ; what was the price of the sixth ? or, in other words, if the first term of a series be 3, the number of terms 6, and the ratio 2, what is the last term ? The first term is 3, the second, 3X2=6, the third, 6X2=13X2X2=) 12, the fourth, 12x2=13X2X2X2=) 24, the fifth, 24X 2=(3x2x2x2x2 =) 48, and the sixth, 48 X2=(3X2X2X 2X2X2=) 96. Ther. 96 cents is the cost of the sixth broom. By examining the above, it will be seen, that the ratio is, in the production of each term of the series, as many times a factor, less one, as the number of terms, and that the first term is always employed once as a factor, or, in other words, any term of a geometrical series is the product of the ratio, raised to a power whose index is one less than the number of the term, multiplied by the first term. NOTE.-If the second power of a number, as 22, be multiplied by the third power, 23, the product is 25. Thus, 22=2 X24, and 23=2x2x28, and 8X4532%2x2x2x2x2; and, generally, the power produced by multiplying one power by another is denoted by the sum of the indices of the given powers. Hence, in finding the higher powers of numbers, we may abridge the operation, by employing as factors several of the lower powers, whose indices added together will make the index of the required power. To find the seventh power of 2, we may multiply the third and fourth powers together, thus : 27=23X24=8X16=128. Ans. I. The first term and ratio given to find any other term. RULE.—Find the power of the ratio, whose index is one less than the number of the required term, and multiply this power by the first term, the product will be the answer, if the series is increasing ; but if it is decreasing, divide the first term by the power. 1. The first term of a ge 2. The first term of a de. ometrical series is 5, the creasing series is 1000, the ratio 3; what is the tenth ratio 4, and the number of term ? terms 5; what is the least 39534X35=81 X 248= term ? 19683, and 19683X5—98415 Ans. Ans. 371 282. A person hought 6 brooms, giving 3 cents for the first, and 96 cents for the last, and the prices form a geometrical series, the ratio of which was 3; what was the cost of all the brooms ? The price would be the sum of the following series : 3-4-6+12+A+ 43+96==189 cents, Ans. If the foregoing scries be multiplied by the ratio 2, the product is 6+12+24+48796=192, whose sum is twice that of the first. Now, subiracting the first series from this, the remainder is 192 3=189=the sum of the first series. Had the ratio been other than 2, the remainder would have been as many times the sum of the series as the ratio, less 1, and the remainder is always the difference between the first term and the product of the last term by the ratio. Hence, II. The first and last term and ratio given to find the sum of the series. RULE.—Multiply the last term by the ratio, and from the product subtract the first term, the remainder divided by the atio, less 1, will give the sum of the series. 2. The first term of a geo- second, $4 the third, and so on, metrical series is 4, the last each succeeding payment beterm 972, and the ratio 3; | ing double the last; and what what is the sum of the series ? | will be the last payment? 3-1)972X3—4=1456 Ans. Ans. $2048 last pay't. NOTE.-The marks drawn over the numbers show, that 4 must be 5. A gentleman, being asktaken from the product of_972, by 3; ed to dispose of a horse, said and the remainder divided by (5-1 =) 2. This mark is called a vincu- | he would sell him on condition luin. of having 1 cent for the first 3. The extremes of a geo- | nail in his shoes, 2 cents for metrical progression are 1024 the second, 4 cents for the and 59049, and the ratio 1.) ; third, and so on, doubling the what is the sum of the series? price of every nail to 32, the Ans. 175099. number of nails in his four 4. What debt will be dis- shoes; what was the price of charged in 12 months, by pay the horse at that rate? ing $1 the first month, $2 the Ans. $42949672.95. 283. If a pension of 100 dollars per annum be forborne 6 years, what is there due at the end of that time, allowing compound interest at 6 per cent. ? Whatever the time, it is obvious that the last year's pension will draw no interest; it is, therefore, only $100; the last but one will draw interest one year, amounting to $106; the last but two, interest (compound) for ? years, amounting to $112,36 ; and so on, forming a geometrical progression, whose first term is 160, the ratio 1.06, and the sum of this series will be the amount due. To find the last term (281), say, 1.065X100= 133.82255776, the sixth term; and to find the sum of the series (282) say, 133.82255777X1.06—100=41.8519112256, which, divided by 1.06–1=0.06, gives $697.5318576 Ans. er sum due. 284. A sum of money payable every year, for a number of years, is called annuity. When the payment of an annuity is forborne, it is said to be in arrears. 1. What is the amount of an an 2. If a yearly rent of $50 be fornuity of $40, to continue 5 years, borne 7 years, to what does it amount, allowing 5 per cent. compound inter per cent, compound interest ? est? Ans. $221.025. Ans. 8394.91. at 4 3. Buodecimals. 285. Of the various subdivisions of a foot, the following is one of the most common: TABLE. 1 foot is 12 inches, or primes, (") = 1 foot. 1 inch “ 12 seconds, '("') 1 second “ 12 thirds, ("") This T= 114 1 third * 12 fourths, (ii) of of 11=1728, &c. forming a decreasing geometrical progression, whose first term is 1, and ratio 12. Hence they are called Duodecimals. 286. How many square feet in a floor, 10ft. 4in. long, und 7ft. Sin. wide ? Here we wish to multiply 10ft. 4! by 10ft. 41 7ft. 8'; we therefore write them as at 7 8 the left hand, and multiply 4 by 8=32; but 4' being is of a foot, and 8' 8, the 6 10 8 product is (148=) of a foot, or 72 4 32'', which reduced gives 2' 81'; put ting down 8'', we reserve the 2 to be 79ft. 2 911 Ans. added to the inches. Multiplying 10ft. by 8=1, the product is (223) 93, to which being added, we have $2=6ft. 10'. Next, multiplying 4 by 7=2=2ft. 4', writing the 4' in the place of inches, and reserving the 2ft., we say 7 times 10 are 70, and two added are 72, which we write under the 6ft., and the sum of these partial products is 79ft. 2' 8Ans. NOTE.-When feet are concerned, the product is of the same denomina. tion as the term multiplying the feet; and when feet are not concerned, the name of the product will be denoted by the sum of the indices of the two factors, or strokes over them. Thus, 4X2"=8" Therefore, 287. To multiply a number consisting of feet, inches, seconds &c. by another of the same kind. RULE.—Write the several terms of the multiplier under the corresponding terms of the multiplicand; then multiply the whole multiplicand by the several terms of the multiplier successively, beginning at the right hand, and placing the first term of each of the partial products under its respective multiplier, remembering to carry one for every 12 from a lower to the next higher denomination, and the sum of these partial products will be the answer, the left hand term being feet, and those towards the right primes, seconds, &c. This is a very useful rule in measuring wood, boards, &c., and for artificers in finding the contents of their work. QUESTIONS FOR PRACTICE. 2. How much wood in a load 8. How many cords in a pile 7ft. 6' long, 4ft. 8' wide, and of 4 foot wood, 24ft. long, and 4ft. high? 6ft. 4' high? Ans. 140ft. or 1 cord 12ft. Ans. 4. cords. Multiply the length by the width, 9. How many square yards and this product by the height. in the wainscoting of a room 3. How many square feet in i8ft. long, 16ft . Ở wide, and a board 16ft. 4in. long, and 2ft. 9ft. 10' high? Sin. wide ? Ans. 75yd. 3ft. 6. Ans. 43ft. Oin. 81. 10. How much wood in a 4. How many feet in a stock cubic pile measuring eft. on of 12 boards 14ft. (' long, and Ans. 4 cords. lft. 3' wide ? Ans. 217ft. 6'. 11. How many square feet Note.—Inches, it will be recol- in a platform, which is 37 feet lected, are so many 12ths of a foot, 11 inches long, and 23 feet 9 whether the foot is lineal, square, inches broad? or solid. in. in the above answer Ans. 900ft. 6' 3". is of a square foot, or 72 square inches. 12. How much wood in a 5. What is the content of a load 8ft. 4in. long, 3ft. Iin. wide, ceiling 43ft. 3' long, and 25ft. and 4ft. 5in. high? @broad? Ans. 138ft. 0' 3". Ans. 1102f. 10' 6". 13. How many feet of floor6. How much wood in a load | ing in a room which is 28fta 6ft. ?! long, 3ft. 5' high, and bin. long, and 23ft. 5ın. broad? 3ft. 8' wide ? Ans. 667ft. 4' 6". Ans. 82ft. 5' 8" 4". 14. How many square feet 7. What is the solid content are there in a board which is of a wall 53ft. 6' long, 12ft. 3 15 feet 10 inches long, and 93 high, and 21t. thick ? inches wide ? Ans. 1310ft. 9', Anis. 12ft. 104" 64 every side ? 4. Position. 288. Position is a rule by which the true answer to a certain class of questions is discovered by the use of false or supposed numbers. 289. Supposing A's age to be clouble that of P's, and B's age triple that of C's, and the sum of their ages to be 140 years; what is the age of each ? Let us suppose C's age to be 8 years, then, by the question, B's age is 3 times 8=24 years, and A's 2 times 24=18, and their sum is (8+24743=) 80. Now, as tie ratios are the same, both in the true and supposed ages, it is evident that the true sum of their ages will have the same ratio to the true age of each individual, that the suin of the supposed ages has to the supposed age of each individual, that is, 30 :8:: 140 : 12, C's true age; or, 80: 24 :: 110:42, B's age, or 30:48 :: 140 : 84, A's age. This operation is called Single Position, and may be expressed as follows: 290. When the result has the same ratio to the supposition that the given number has to the required one. RULE.-Suppose a number, and perform with it the operation described in the question. Then, by proportion, as the result of the operation is to the supposed number, so is the given result to the true number required. 2. What number is that, 4. A vessel has 3 cocks ; which, being increased by }, the first will fill it in 1 hour, and itself, will be 125? the second in 2, the third in Then 50 : 24 : : 125 : C0 Ans. 3; in what time will they all Sup. 24 Or by fractions. fill it together? Ans. hour. 1 and 1 of his money, had 1+1+1+=125, $60 left; what had he at Result 50 or istitut first ? Ans. $144. I'=41, and 1= 6. What number is that, 2 ) 125 ( 60 Ans. from which, if 5 be subtract(See p. 104, Miscel.) ed, of the remainder will 3. What number is that be 40 ? Ans. 65. whose 6th part exceeds its 8th part by 20? Ans. 480. II. When the ratio between the required and the supposed number differs from that of the given number to the required one. 291. RULE.—Take any two numbers, and proceed with each according to the condition of the question, noting the |