591. To find the side of a cube whose solidity shall be dou ble, triple, &c., that of a cube whose side is given. Cube the given side, multiply it by the given proportion, and the cube root of the product will be the side of the cube required. 23. What is the side of a cubical bin which contains 8 times as many solid feet as one whose side is 4 feet? Ans. 8 ft. 24. What is the side of a cubical block which contains 4 times as many solid yards as one whose side is 6 feet ? 25. If a ball 6 inches in diameter weighs 32 lbs., what is the weight of a ball whose diameter is 3 inches? 26. If a globe 4 ft. in diameter weighs 900 lbs., what is the weight of a globe 3 ft. in diameter ? 592. To find two mean proportionals between two given numbers. Divide the greater number by the less, and extract the cube root of the quotient. Multiply the root thus found by the least of the given numbers, and the product will be the least proportional sought; then multiply the least mean proportional by the same root, and this product will be the greater mean proportional required. Find two mean proportionals between the following numbers: 27. 8 and 216. 29. 12 and 1500. 31. 71 and 15336. 28. 64 and 512. 30. 40 and 2560. 32. 83 and 60507. EXTRACTION OF ROOTS OF HIGHER ORDERS. 593. When the index denoting the root to be extracted is a composite number. First extract the root denoted by one of the prime factors of the given index; then of this root extract the root dénoted by another prime factor, and so on. Thus, For the 4th root, extract the square root twice. For the 6th root, extract the cube root of the square root. Ans. 3. 2. What is the 8th root of 256 ? 3. The 4th root of 65536 ? 4. The 4th root of 19987173376 ? 5. The 6th root of 46656 ? 8. The 9th root of 40353607 ? 10. The 27th root of 134217728 ? 594. When the index denoting the root is not a composite number, we have the following general RULE FOR EXTRACTING ALL ROOTS. I. Point off the number into periods of as many figures each, as there are units in the given index, commencing with the units figure. II. Find the first figure of the root, and subtract its power from the left hand period; then to the right of the remainder bring down the first figure in the next period for a dividend. III. Involve the root to the power next inferior to that of the index of the required root, and multiply it by the index itself, for a divisor. IV. Find how many times the divisor is contained in the dividend, and the quotient will be the next figure of the root. V. Involve the whole root to the power denoted by the index of the required root, and subtract it from the two left hand periods of the given number. VI. Finally, bring down the first figure of the next period to the remainder, for a new dividend, and find a new divisor as before. Thus proceed till the whole root is extracted. OBS. 1. The reason of this rule may be illustrated in the same manner as that for the extraction of the Square and Cube Roots. 2. The proof of all roots is by involution. 3. Any root whatever may be extracted by an extension of the principle applied to the extraction of the cube root. In this general application of the principle, the given number must be divided into periods, each consisting of as many figures as there are units in the index of the required root, and the number of columns employed will be one less than there are units in the given index. The operation then proceeds exactly as in the extraction of the cube root; and if there be a remainder, a like contraction is admissible. 12. Required the 5th root of 95†. 13. Required the 7th root of 2103580000000000000. Note. The preceding method in most of the practical cases, gives perhaps as easy solutions, as the nature of the case will admit. But when roots of a very high order are required, the process may be shortened by the following * root. APPROXIMATE RULE. 595. Call the index of the given power n; and find by trial a number nearly equal to the required root, and call it the assumed Raise the assumed root to the power whose index is n. Then, As n+1 times this power, added to n―1 times the given number, is to n―1 times the same power added to n+1 times the given number, so is the assumed root to the true root nearly. The number thus found may be employed as a new assumed root, and the operation repeated to find a result still nearer the true root. 14. Required the 365th root of 1.06. Solution.-Take 1 for the assumed root, the 365th power of which is 1; and n being 365, we have n+1=366, and n—1= Then proceed in the following manner: 364. SECTION XVIII PROGRESSION. ART. 596. When there is a series of numbers such, that the ratios of the first to the second, of the second to the third, &c., are all equal, the numbers are said to be in Continued Proportion, or Progression. Progression is commonly divided into arithmetical and geometrical. Note. The terms arithmetical and geometrical are used simply to distinguish the different kinds of progression. They both belong equally to arithmetic and geometry. ARITHMETICAL PROGRESSION. 597. Numbers which increase or decrease by a common difference, are in arithmetical progression. (Art. 474. Obs.) OBS. 1. Arithmetical progression is sometimes called progression by difference, or equidifferent series. 2. When the numbers increase, the series is called ascending; as, 3, 5, 7, 9, 11, &c. When they decrease, the series is called descending; as, 11,9,7, 5, &c. 598. When four numbers are in arithmetical progression the sum of the extremes is equal to the sum of the means. Thus, if 5-3-9-7, then will 5+7=3+9. Again, if three numbers are in arithmetical progression, the sum of the extremes is double the mean. Thus, if 9-6-6-3, then will 9+3=6+6. 599. In any arithmetical progression, the sum of the two extremes is equal to the sum of any other two terms equally distant from the extremes, or equal to double the middle term, when the number of terms is odd. Thus, in the series 1, 3, 5, 7, 9, it is obvious that 1+9=3+7=5+5. 600. In an ascending series, each succeeding term is found by adding the common difference to the preceding term. Thus, if the first term is 3, and the common difference 2, the series is 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, &c. In a descending series, each succeeding term is found by subtracting the common difference from the preceding term. Thus, if the first term is 15, and the common difference 2, the series is 15, 13, 11, 9, 7, &c. 601. In arithmetical progression there are five parts to be considered, viz: the first term, the last term, the number of terms, the common difference, and the sum of all the terms. These parts have such a relation to each other, that if any three of them are given, the other two may be easily found. 602. If the sum of the two extremes of an arithmetical progression is multiplied by the number of the terms, the product will be double the sum of all the terms in the series. Take the series The same inverted The sums of the terms are 2, 4, 6, 8, 10, 12. 12, 10, 8, 6, 4, 2. 14, 14, 14, 14, 14, 14. Thus, the sum of all the terms in the double series, is equal to the sum of the extremes repeated as many times as there are terms; that is, the sum of the double series is equal to 12+2 multiplied by 6. But this is twice the sum of the single series. Hence, 603. To find the sum of all the terms, when the extremes and the number of terms are given. Multiply half the sum of the extremes by the number of terms, and the product will be the sum of the given series. OBS. The reason of this process is manifest from the preceding illustration. Ex. 1. The extremes of a series are 3 and 25, and the number of terms is 12: what is the sum of all the terms? Ans. 168. 2. What is the sum of the natural series of numbers, 1, 2, 3, 4, 5, &c., up to 100? 3. How many strokes does a common clock strike in 12 hours? 604. To find the common difference, when the extremes and the number of terms are given. Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference required. OBS. The truth of this rule is manifest from Art. 602. |