sum; subtract 1 from the index, and call the result, the diffe rence. IV. Multiply the power by the sum, and the number by the difference, and add both products together for the first term. V. Multiply the number by the sum, and the power by the difference, and add both products together for the second term. VI. Make a rule of three stating, thus; say as the first term: is to the second term :: so is the assumed root: to a fourth number, (found by the rule of three,) which will be the root of the given number nearly. VII. Involve the root found to the given power, and if the power and given number are nearly equal, the work is finished; but if not, the operation must be repeated, thus ; VIII. Let the root found be called the assumed root, and its power the power, and proceed with these and the given number, sum, and difference, as before, whence a root will be obtained still nearer the truth. In this manner the operation may be repeated at pleasure, observing always to use the last found root, and its power, for the assumed root and power. EXAMPLES. 1. Required the cube root of 520. Here 520 the number. 3= the index. 3+1=4= the sum. 3-1=2 = the difference. I find by trials that 8 is nearly equal to the cube root of 520; therefore 8 = the assumed root, and 8)3 = 512 = the power. Then 512 × 4 = 2048 = the power multiplied by the sum. 520 × 2 = 1040 = the number multiplied by the diffe their sum = 3088 the first term. [rence. And 520 × 4 = 2080 = the number multiplied by the sum. 512 × 2 = 1024 = the power multiplied by the diffe their sum 3104 = the second term. [rence. 2. Required the 5th root of 40. Here 40 the number. 5 sum. 5—1=4 = the difference. the index. 5+ 16 = the Let the root found by trials be 2 = the assumed root; then Ã3 = 32 = the power. Here 2.0909 = the root assumed. 2.0909 = 39.963757, &c. 4 x 39.963757 = 159.855029 Second term 399.855029 Wherefore 399.782543: 399.855029 :: 2.0909 : 4. What is the 4th root of 94.75853? Root 3.1196, &c. 5. What is the 5th root of 3124? Root 4.9996, &c. 6. Extract the 6th root of 48. Root 1.90636, &c. 7. Required the 7th root of 581. Root 2.4824, &c. 8. Required the 8th root of 72138957.88. Root 9.5999, &c. 9. What is the 9th root of 2 ? Root 1.080059, &c. 10. Extract the second, third, fourth, fifth, sixth, seventh, eighth, and ninth roots of one hundred, and find their sum. Ans. 27.84716, &c. 284. In all the foregoing examples the index of the root is a fraction, having 1 for its numerator; examples however sometimes oceur, in which the numerator of the index is greater than 1; in this case the root is extracted by the following: RULE I. Involve the given number to that power which is denoted by the numerator of the index, from whence extract the root denoted by the denominator: or, II. First extract the root denoted by the denominator, then involve this root to the power denoted by the numerator f. 11. Find the value of 84. 3 Thus by Rule I. 92=64, and 3/644, the root required. By Rule II. 3/8±2, and 2}2 = 4, the root as before. 3 12. Required the value of 10. Thus 10+= 10000 and 5/10000 = 6.3096, &c. the root: or 5/10 = 1.5849, &c. and 1.5849)* = 6.3096, the root as before. 13. Required the value of 4. Ans. 1.68179, &c. 14. Required the value of 1023. Ans. 15,993, &c. 15. Required the value of 10648}}. Ans. 484. 16. Required the sum of the values of 4, 34, 4, and 3). Ans. 10.72196, &c. PROGRESSION. 285. When several numbers or terms are placed in regular succession, the whole is called a series. 286. If the terms of a series successively increase or decrease, according to some given law, the series is said to be in progression. 287. Progression is of two kinds, Arithmetical and Geometri f In this rule both Involution and Evolution are employed; the numerator of the fractional index denoting a power, and the denominator a root; thus in ex. 11. 8 denotes either the cube root of the square of 8, or the square of the cube root of 8: on this principle the rule depends. cal, arising from the manner in which the successive increase or decrease is made; namely, either by addition or subtraction, or by multiplication or division. ARITHMETICAL PROGRESSION. 288. A series of numbers is said to be in Arithmetical Progression, when the terms successively increase or decrease by the constant addition or subtraction of a number, called the common difference . There are five particulars belonging to questions in arithme tical progression; viz. 1. The least term, 2. The greatest term, } called the extremes. 3. The number of terms. 4. The common difference. 5. The sum of all the terms. Any three of these five being given, the remaining two may be found, as is shewn by the rules and examples following. 289. The least term, the greatest term, and the number of terms, being given, to find the sum of all the terms. RULE. Add the least and greatest terms together, multiply the sum by half the number of terms, and the product will be the sum required. EXAMPLES. 1. The least term is 3, the greatest 17, and the number of terms 8, in an arithmetical progression; required the sum of the terms. Thus 3 + 17 = 20 = sum of the extremes. And 4 (or half 8) = half the number of terms. 8 When the progression consists of three or four terms only, it is usually called an arithmetical proportion; and the middle terms are called arithme tical means. The essential property of an arithmetical progression is this; namely, "The sum of the two extreme terms is equal to the sum of every two mean "terms equally distant from the extremes;" from this property many others, some of which are the subject of the following rules, are easily deduced; but as this cannot be conveniently done without Algebra, it was thought best to refer to the Algebraic part of the work for proof of the rules here given. The word progression is derived from the Latin progredior, to go forward. 2. The least term is 5, the greatest 205, and the number of terms 11, being given, to find the sum of the terms. Thus 5 + 205 = 210 = sum of the extremes. Wherefore 210 × 5+ = 1155 = the sum required. 3. The extremes are 4 and 800, and the number of terms 40, to find the sum. Thus 4+ 800 × 20 = 16080, the sum required. 4. A man paid a debt which he owed at 20 payments in arithmetical progression; the first payment was 31. and the last 187. what was the debt? Ans. 210l. 5. I bought 100 peaches, and paid for them in arithmetical progression, viz. for the first d. and for the last 6d. what sum did the whole amount to? Ans. 11. 7s. 1d. 6. What must be given for 120 elm trees, the prices whereof are in arithmetical progression, that of the first being 5s. and that of the last 101. Ans. 615l, 290. The least term, the greatest term, and the number of terms, being given, to find the common difference. RULE. Subtract the least term from the greatest, and divide the remainder by 1 less than the number of terms; the quotient will be the common difference required. 7. In an arithmetical progression, the least term is 3, the greatest 17, and the number of terms 8; required the common difference? Thus 17-3 = 14= the difference of the extremes. And 8-17= the number lessened by 1. 14 7 Therefore =2= the common difference sought. 8. The least term is 5, the greatest 205, and the number of terms 11, in an arithmetical progression; required the common difference? 205-5 Thus 11-1 200 =20= the common difference required. 9. A man had 5 sons, whose ages were in arithmetical progression, the youngest was 3 years old, when the eldest was 13'; required the common difference of their ages? |