6. What is the cube root of .000097336 ? 8 10. What is the cube root of .015625 ? Ans. .046. 'Ans. 5. Ans. 2.42+. Ans. .25. ARITHMETICAL PROGRESSION. ART. 171, When several numbers are so arranged as to increase or decrease in regular order by a common difference, they are said to be in arithmetical progression. When they increase by the addition of a constant number, it is called an ascending series, e. g., 1, 3, 5, 7, 9, 11, 13, &c. When they decrease by the subtraction of a constant number, it is called a descending series, e. g., 19, 16, 13, 10, &c. The numbers are called terms, the first and last being called extremes, and the intermediate terms the means. In arithmetical progression there are five quantities so related to each other, that any three of them being given, the remaining two may be found. This fact gives rise to twenty different cases or problems, only six of which will here be given. These five quantities in the formulas expressing their relation, are represented as follows: (1), a, d and n being given, l=a±(n−1)d, The interpretation of these formulas for those not familiar with algebraic expressions, will furnish the following rules. The student will be able to select the proper rule for any particular case by noting carefully which three of the five quantities are given, and which is required. (1). The first term, common difference, and number of terms being given to find the last term. RULE.-Multiply the common difference by the number of terms, less one, and add the product to the first term, if the series be ASCENDING, but subtract it from it, if the series be DESCENDING. (2). The first term, number of terms, and last term being given to find the common difference. RULE.-Divide the difference of the extremes by the number of terms, less one. (3). The first term, common difference, and last term being given to find the number of terms. RULE.-Divide the difference of the extremes by the common difference, and add 1 to the quotient. (4). The first term, number of terms, and last term being given to find the sum of all the terms. RULE.-Multiply half the sum of the extremes by the number of terms. (5). The common difference, number of terms, and sum of all the terms being given to find the first term. RULE. Divide the sum of the terms by the number of terms; subtract from the quotient, if the series be ascending, otherwise add to it half the product of the common difference into the number of terms, less one. (6). The first term, common difference, and number of terms being given to find the sum of all the terms. RULE. Add to twice the first term, if the series be ascending; otherwise subtract from it the product of the common difference into the number of terms, less one; multiply the sum or difference by half the number of terms. Examples. 1. A laborer agreed to dig a well 100 feet deep, for which he was to receive 1 cent for the first foot, 5 cents for the second and so on increasing the price 4 cents per foot for the entire depth. What would he get for the last foot? Ans. $3.97. 2. If a man begin by lifting 200 lbs., and make equal additions to the weight daily for a year of 365 days, what must be the daily additions to reach 800 lbs. at the end of the year? Ans. 1 lbs. Secondly. With what weight must he begin, so that the daily additions may be two pounds? Ans. 72 lbs. Thirdly. If he begin with 200 lbs., and add 1 lbs. daily, how many days would it require to reach 800 lbs. Ans. 401 days. Ans. 78. 3. How many strokes does the hammer of a clock make in 12 hours ? 4. If 100 stakes be set in a straight line 10 feet apart, how much twine will it require to connect the first one in the line with each of the others separately? Ans. 49500 feet. 5. A man agreed to contribute for a benevolent object one cent the first day, two cents the second day, three cents the third day, and so on through the year of 365 days. What was the amount of his donation ? Ans. $667.95. 6. A note was given for $1000, with interest payable annually, at 7%. Nothing having been paid for ten years, how much did the total amount of interest due exceed the simple interest of the principal? See Art. 100. Ans. $220.50. 7. If a note for $2000, drawing interest at 6% per annum, run 10 yrs. 3 mo. 9 d. with nothing paid, how much would the condition of making the "interest payable semi-annually" increase the amount due ? Ans. $361.80. GEOMETRICAL PROGRESSION. ART. 172. A geometrical progression is such a series of numbers, that each term after the first shall be the product of the preceding term and a constant multiplier, called the common ratio. The progression is ascending or descending, according as the ratio is greater or less than unity. In geometrical progression, as in arithmetical progression, there are five quantities so related to each other, that any three of them being given the remaining two may be found. Of the twenty cases arising therefrom only four will here be noticed. In the formulas expressing the relation of the five quantities referred to above, they are represented as follows: The first term. a These formulas are equivalent to the following rules: (1.) The first term, common ratio, and number of terms being given to find the last term. RULE.-Raise the common ratio to a power whose degree is one less than the number of terms, and multiply it by the first term. (2.) The last term, common ratio, and number of terms being given to find the first term. RULE.-Raise the common ratio to a power whose degree is one less than the number of terms, and divide the last term by it. (3.) The first term, last term, and common ratio being given to find the sum of all the terms. RULE. From the product of the last term into the ratio, subtract the first term; then divide the remainder by the ratio less one. (4) The first term, common ratio, and number of terms being given to find the sum of all the terms. RULE. From the power of the ratio whose degree is the number of terms, subtract one; divide the remainder by the common ratio less one, and multiply the quotient by the first term. REMARK. It is sometimes convenient in working problems to transpose a descending series so as to make it ascending, the last term of the first series becoming the first term in the new. In that case the new ratio would be the reciprocal of the old, i. e., unity divided by that ratio, e. g., would become 3, would become 4, and so on. Infinite Series. ART. 173. If the number of terms in a descending geometrical series be infinite, the last term will be 0. It does not, however, follow that, because the number of terms is infinite, the sum of those terms must be infinite, for if we apply formula (3) making the last term 0, we shall find the sum of an infinite decreasing series to be the first term divided by the difference between the common ratio and unity, 1. A man offered to purchase 10 cows, paying for the first 5 cents, for the second 15 cents, and so on tripling the amount for each succeeding cow. What would the last one cost him, and what would the whole cost him? Ans. The last would cost $984.15. |