Now, by observing this operation, and remarking what would be lost in the course of it by omitting the second figure of the root, 6, taking 20 instead of 26, we see that when we have found the 20 the next inquiry is, what number must be added to 20, so that, if we multiply it into itself once and into 20 twice, and the sum of these products, together with the second power of twenty, by twenty plus this number, the result will be 17576, or 8000 + 9576. Then, in order to obtain this number, or an approximation to it, we take twice the part of the root found, 20, and multiply the result, 40, by 20, as we should do in raising it to the third power, and make this, which is 800, a part of the divisor, and the product of which by 6 was lost in the operation for want of the 6 added to 20. But this is not all the loss. There is also the second power of 20 by 6, and therefore 400 to be added to the 800 for a divisor. There still remains the further loss of the third power of 6 (216), and also of 6 times 240 and 20 times 36 ; but these we neglect in the formation of the divisor. The divisor is contained in the dividend 7 times; but, making the allowance of a unit for the neglect of the numbers above named, we take 6 for the quotient figure, and proceed to find the subtrahend, which, according to the rule and the foregoing operation of raising 20 + 6 to the third power, must be 1200 X 6 + 1440 + 720+ 216 = 17576. APPLICATION OF THE CUBE ROOT. PRINCIPLES ASSUMED. Spheres are to each other as the cubes of their diameter. Cubes, and all solids whose corresponding parts are similar. and proportional to each other, are to each other as the cubes of their diameters, or of their homologous sides. 29. If a ball, 3 inches in diameter, weigh 4 pounds, what will be the weight of a ball that is 6 inches in diameter? Ans. 32lbs. 30. If a globe of gold, one inch in diameter, be worth $120, what is the value of a globe 3.1 inches in diameter ? Ans. $ 5145. 31. If the weight of a well-proportioned man, 5 feet 10 inches in height, be 180 pounds, what must have been the weight of Goliath of Gath, who was 10 feet 4 inches in height? Ans. 1015.1+ lbs. 32. If a bell, 4 inches in height, 3 inches in width, and of an inch in thickness, weigh 2 pounds, what should be the dimensions of a bell that would weigh 2000 pounds ? Ans. 3ft. 4in. high, 2ft. 6in. wide, and 2 in. thick. 33. Having a small stack of hay, 5 feet in height, weighing lcwt., I wish to know the weight of a similar stack that is 20 feet in height. Ans. 64cwt. 34. If a man dig a small square cellar, which will measure 6 feet each way, in one day, how long would it take him to dig a similar one that measured 10 feet each way ? Ans. 4.629+ days. 35. If an ox, whose girth is 6 feet, weighs 600lbs., what is the weight of an ox whose girth is 8 feet ? Ans. 1422.2+Ibs. 36. Four women own a ball of butter, 5 inches in diameter. It is agreed that each shall take her share separately from the surface of the ball. How many inches of its diameter shall each take? Ans. First, .45+ inches ; second, .57+ inches ; third, .82+ inches; fourth, 3.149+- inches. 37. John Jones has a stack of hay in the form of a pyramid. It is 16 feet in height, and 12 feet wide at its base. It contains 5 tons of hay, worth $ 17.50 per ton. Mr. Jones has sold this hay to Messrs. Pierce, Rowe, Wells, and Northend. As the upper part of the stack has been injured, it is agreed that Mr. Pierce, who takes the upper part, shall have 10 per cent. more of the hay than Mr. Rowe; and Mr. Rowe, who takes his share next, shall have 8 per cent. more than Mr. Wells; and Mr. Northend, who has the bottom of the stack, that has been much injured, shall have 10 per cent. more than Mr. Wells. Required the quantity of hay, and how many feet of the height of the stack, beginning at the top, each receives. Ans. Pierce receives 274cwt. and 10.366+ feet in height; Rowe, 2419|cwt. and 2.493 feet ; Wells, 2234fcwt. and 1.666 feet; Northend, 25,5cwt. and 1.474 feet. A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. Rule. — 1. Prepare the given number for extraction, by pointing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, or by inspection, in the table of powers, and subtraci its power from the left-hand period. 3. To the remainder, bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferior power to that which is given, and multiply it by the number denoting the given power for a divisor. 5. Find how many times the divisor is contained in the dividend, and the quotient will be another figure of the root. An allowance of two or three units is generally made, and for the higher powers a still greater allowance is necessary. 6. Involve the whole root to the given power, and subtract it from the given number, as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, as before; and in like manner proceed till the whole is finished. 1. What is the cube root of 20346417? OPERATION. 23 = 20346417(273 = lst subtrahend. 19683 - 2d subtrahend. 272 X 3=2187) 6634 = 2d dividend 20346417 = 3d subtrahend. 2733 = 2. What is the fourth root of 34828517376 ? OPERATION. 44 = 34828517376(432 Ans. = Ist subtrahend. 3418801 -= 2d subtrahend. 433 X 4 = 318028) 640507 = 2d dividend. 34828517376 = 3d subtrahend. 4324 = 3. What is the 5th root of 281950621875 ? Ans. 195. 4. Required the sixth root of 1178420166015625. Ans. 325. 5. Required the seventh root of 1283918464548864. Ans. 144. 6. Required the eighth root of 218340105584896. Ans. 62. SECTION LXIII. ARITHMETICAL PROGRESSION. WHEN a series of antities or numbers increases or de. creases by a constant difference, it is called arithmetical progression, or progression by difference. The constant difference is called the common difference or the ratio of the progression. Thus, let there be the two following series : 1, 5, 9, 13, 17, 21, 25, 29, 33, 25, 22, 19, 16, 13, 10, 7, 4, 1. The first is called an ascending series or progression. The second is called a descending series or progression. The numbers which form the series are called the terms of the progression. The first and last terms of the progression are called the extremes, and the other terms, the means. Any three of the five following things being given, the other two may be found : 1st, the first term, PROBLEM I. The first term, last term, and number of terms being given, to find the common difference. To illustrate this problem, we will examine the following series, 3, 5, 7, 9, 11, 13, 15, 17, 19. It will be perceived that in this series 3 and 19 are the extremes, 2 the common difference, 9 the number of terms, and 99 the sum of the series. It is evident, that the number of common differences in any number of terms will be one less than the number of terms. Hence, if there be 9 terms, the number of common differences will be 8, and the sum of these common differences will be equal to the difference of the extremes; therefore if the difference of the extremes, 19. 16, be divided by the number of common differences, the quotient will be the common differ Thus 16 = 8= 2 is the common difference. 3 = ence. Rule. — Divide the difference of the extremes by the number of terms less one, and the quotient is the common difference. 1. The extremes are 3 and 45, and the number of terms is 22. What is the common difference? OPERATION. 45-3 2 Answer. 22 1 2. A man is to travel from Albany to a certain place in 11 days, and to go but 5 miles the first day, increasing the distance equally each day, so that the last day's journey may be 45 miles. Required the daily increase. Ans. 4 miles. 3. A man had 10 sons, whose several ages differed alike; the youngest was 3 years, and the oldest 48. What was the common difference of their ages? Ans. 5 years. 4. A certain school consists of 19 scholars; the youngest is 3 years old, and the oldest 39. What is the common difference of their ages ? Ans. 2 years. PROBLEM II. The first term, last term, and number of terms being given, to find the sum of all the terms. Illustration. Let 3, 5, 7, 9, 11, 13, 15, 17, 19, be the series, and 19, 17, 15, 13, 11, 9, 7, 5, 3, the same series inverted. 22, 22, 22, 22, 22, 22, 22, 22, 22, sum of both series. From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series. It is evident that if, in the above series, 22 be multiplied by 9, the number of terms, the product will be the sum of both series, 22 X 9 = 198, and therefore the sum of either series will be 198 ; 2= 99. But 22 is also the sum of the extremes in either series, 3 + 19 = 22. Therefore, if the sum of the extremes be multiplied by the number of terms, the product will be double the sum of the series. Rule Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series. 5. The extremes of an arithmetical series are 3 and 45, the number of terms 22. Required the sum of the series. and |