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, 3, &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 third, 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, 3×2-6, the third, 6×2 (3×2×2=) 12, the fourth, 12×2=(3×2×2×2=) 24, the fifth, 24×2 (3×2×2×2×2 =) 48, and the sixth, 48×2=(3×2×2×2×2×2=) 96. Then 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 X2=4, and 23=2×2×2=8, and 8×4=32—2×2×2×2×2; 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 8×16-128. 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 geometrical series is 5, the ratio 3; what is the tenth term? 39-34X35 81X248 19683, and 19683X5-98415 Ans. 2. The first term of a decreasing series is 1000, the ratio 4, and the number of terms 5; what is the least term? Ans. 3. 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: 3-1-6+12+24+ 48+96-189 cents, Ans. If the foregoing series be multiplied by the ratio 2, the product is 6+12+24+48+96=192, whose sum is twice that of the first. Now, subtracting 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- | metrical series is 4, the last term 972, and the ratio 3; what is the sum of the series? | =) 2. This mark is called a vinculum. second, $4 the third, and so on, each succeeding payment being double the last; and what will be the last payment? $4095 the debt. Ans. { $2048 last pay't. 5. A gentleman, being ask 3-1)972×3-4(=1456 Ans. NOTE. The marks drawn over the numbers show, that 4 must be taken from the product of 972, by 3, and the remainder divided by (5-1ed to dispose of a horse, said he would sell him on condition of having 1 cent for the first nail in his shoes, 2 cents for the second, 4 cents for the third, and so on, doubling the price of every nail to 32, the number of nails in his four shoes; what was the price of the horse at that rate? 3. The extremes of a geometrical progression are 1024 and 59049, and the ratio 1; what is the sum of the series? Ans. 175099. 4. What debt will be discharged in 12 months, by paying $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 2 years, amounting to $112,36; and so on, forming a geometrical progression, whose first term is 100, 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, 133.82255777×1.06-100-41.8519112256, which, divided by 1.06-10.06, 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 annuity of $40, to continue 5 years, allowing 5 per cent. compound interest? Ans. $221.025. 2. If a yearly rent of $50 be forborne 7 years, to what does it amount, at 4 per cent. compound interest? Ans. $394.91. 3. Duodecimals. 285. Of the various subdivisions of a foot, the following is one of the most common: 1 foot 1 inch TABLE. is 12 inches, or primes, (') 1= 12 thirds, ("") 12 fourths, () 1 third of of 178, &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, and 7ft. 8in. wide? 10ft. 4/ 7 8 6 10 8 72 4 Here we wish to multiply 10ft. 4' by 7ft. 8'; we therefore write them as at the left hand, and multiply 4 by 8=32; but 4' being of a foot, and 8' 1, the product is (X) of a foot, or 32, which reduced gives 2' 8'; putting down 8, we reserve the 2' to be 79ft. 2/ S" Ans. added to the inches. Multiplying 10ft. by 8', the product is (223) 82, to which being added, we have 36ft. 10. Next, multiplying 4' by 7=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′ 8′′ Ans. NOTE. When feet are concerned, the product is of the same denomination 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, 4×2′′=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 7ft. 6' long, 4ft. 8′ wide, and 4ft. high? Ans. 140ft. or 1 cord 12ft. Multiply the length by the width, and this product by the height. 3. How many square feet in a board 16ft. 4in. long, and 2ft. Sin. wide? Ans. 43ft. 6in. 8". 8. How many cords in a pile Ans. 4 cords. Ans. 75yd. 3ft. 6'. 4. How many feet in a stock cubic pile measuring 8ft. on of 12 boards 14ft. 6' long, and lft. 3' wide? Ans. 217ft. 6'. NOTE.-Inches, it will be recollected, are so many 12ths of a foot, whether the foot is lineal, square, or solid. Gin. in the above answer is a square foot, or 72 square inches. 5. What is the content of a ceiling 43ft. 3′ long, and 25ft. & broad? Ans. 1102ft. 10′ 6′′. every side? Ans. 4 cords. 11. How many square feet in a platform, which is 37 feet 11 inches long, and 23 feet 9 inches broad? Ans. 900ft. 6' 3". 12. How much wood in a load 8ft. 4in. long, 3ft. 9in. wide, and 4ft. 5in. high? Ans. 138ft. 0′ 3′′. 13. How many feet of floor 6. How much wood in a loading in a room which is 28ft. 6ft. 7 long, 3ft. 5′ high, and 3ft. 8' wide? Ans. 82ft. 5' 8" 4"". 7. What is the solid content of a wall 53ft. 6' long, 12ft. 3' high, and 2ft. thick? Ans. 1310ft. 9'. 6in. long, and 23ft. 5in. broad? 14. How many square feet Ans. 12ft. 10 4" 6" 1 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 double that of E'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 & years, then, by the question, B's age is 3 times 8-24 years, and A's 2 times 24-48, and their sum is (8+24+48=) 80. Now, as the 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 sum of the supposed ages has to the supposed age of each individual, that is, 80: 8:: 140: 12, C's true age; or, 80: 24: 140:42, B's age, or 80: 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, which, being increased by, and itself, will be 125? Then 50 : 24 : : 125: C0 Ans. Sup. 24 Or by fractions. Let 1 denote the required number: =12 Result 50 then 4. A vessel has 3 cocks; the first will fill it in 1 hour, the second in 2, the third in 3; in what time will they all fill it together? Ans. hour. 5. A person, after spending and of his money, had 1+1+1+1=125, $60 left; what had he at first? or 12 +++ 5)125 (60 Ans. (See p. 104, Miscel.) 3. What number is that whose 6th part exceeds its 8th part by 20? Ans. 480. Ans. $144. 6. What number is that, from which, if 5 be subtract 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 |