he harvest in the 15th year, the annual increase being 12 fold? 12. A farmer inquiring the price of a drove of 30 oxen, was told that he might have the whole drove for the price of the 20th ox, valuing the first at one cent, the second at 2 cents, the third at 4 cents, and so on, doubling the price of each ox for the price of the next. What would be the price per head at that rate? 13. What is the amount of $275 for 9 years at 4 per cent. compound interest? 14. What sum would amount to $300 in 10 years at 5 per cent. compound interest? 15. What sum of money would amount to $1000 in 12 years at 6 per cent, compound interest? 16. What sum would amount to $2500 in 3 years at 8 per cent. compound interest? 17. What would be the amount of $2300 in 13 years at 7 per cent. compound interest? 18. The first term of a geometrical series is 4194304, and the ratio . What is the fourteenth term? PROBLEM II. The extremes and ratio being given, to find the sum of the terms. The first term of a series is 162, the last term is 2, and the ratio }. What is the sum of the series? The series is 162, 54, 18, 6, 2, the sum of which is 242. 54, 18, 6, 2, 3, is another series obtained by multiplying each term of the first series by the ratio. Subtracting the second series from the first, we have 1611, which is the difference between the first term and the last term multiplied by the ratio. It is also of the sum of the first series, since it is obtained by subtracting of the series from the whole series. Then if we divide by 3, or the difference between the ratio and 1, we shall obtain the desired sum. RULE. Multiply the last term by the ratio, and divide the difference between the product and the first term by the difference between the ratio and 1. 19. The first term 9, the ratio, and the last term 23, are given. What is the sum of the series? 20. What is the sum of 12 terms of the progression 2, 8, 32, &c.? (The last term is found by Problem I.) 21. What is the sum of the series 2, 1, 1, 1, &c. to infinity? (The last term in any infinite decreasing series is 0.) 22. Required the sum of the infinite series 7, 3, 3, &c.? 23. If I lay up $100 every year, to what will the whole amount in 10 years, at 6 per cent. compound interest? (The 1st term is 100, the ratio 1.06, and the number of terms 10.) 24. What would be the amount of an annual saving of $300 for thirty years, at 6 per cent. compound interest? 25. Sysla, the reputed inventor of the game of chess, is said to have asked as a reward, one grain of wheat for the first square on the chess-board, two for the second, and so on in geometrical progression. What would have been the amount of his reward, there being 64 squares on the board, and 9200 grains of wheat in a pint? What would be the height of a cubical bin that would contain it, supposing the base to be 10 miles square? 26. A blacksmith agreed to shoe a horse for the amount of 32 nails, at 1 mill for the first nail, 2 mills for the second, 4 for the third, and so on. What was his charge? 27. What is the sum of the infinite series 6, 1, §, 332, 3, &c.? 28. What is the sum of 20 terms of the series 10, 5, 21⁄21⁄2, 14, &c.? 29. If a man commences at 21 years of age, and annually puts $500 at compound interest, how much will he be worth when he is 50 years old? PROBLEM III. The extremes and ratio being given, to find the number of terms. RULE. Divide the last term by the first, and add one to the index of that power of the ratio, which is equivalent to the quotient. 30. A man made several payments in Geometrical Progression, each being twice as large as the preceding. The first payment was $4, and the last $512. What was the number of payments, and what was the whole amount of the debt? PROBLEM IV. The extremes and number of terms being given, to find the ratio. RULE. Divide the last term by the first, and extract the root of the quotient which is indicated by the number of terms less one. This and the preceding rule, are readily deduced from Problem I. 31. The first term in a Geometrical series is 1, and the eleventh term is 1024. What is the ratio? 32. The first term is 256, and the fifth term 81. What is the ratio? 33. The first ber of terms 6. term is 2, the last term 64 and the numWhat is the ratio? 31259 34. An estate of $200 amounted to $264.86, in seven years at compound interest. What was the rate per cent? 35. At what rate per cent. will $1000 amount to $1689.48, in nine years at compound interest? 36. Insert 2 mean proportionals between 1 and 6. (As there are to be 2 means, the number of terms is 4, and the extremes 1 and 6.) 37. Insert 4 mean proportionals between 3 and 96. 38. Insert 5 mean proportionals between 4 and 2916. 39. Insert 3 mean proportionals between 5 and 6480. 40. Insert 4 mean proportionals between 1 and 4. The remaining Problems in Geometrical Progression, are of little interest. By observing the formulas on the following page, we shall perceive a striking analogy between Arithmetical and Geometrical Progression. The continual product of all the terms in a Geometrical series, is denoted by p. (a l)n n 1 (a + 1) n 2 p = Thus in comparing the first table with the second, we see that multiplication corresponds to addition; 66 "subtraction; division "multiplication; 66 If, therefore, we had a series of numbers bearing the same ratio to the natural series, as an Arithmetical to a Geometrical Progression, the labor of multiplication would be reduced to that of simple addition, and involution to simple multiplication. Such a series constitutes a TABLE OF LOGA RITHMS. CHAPTER XIX. HARMONICAL PROGRESSION.* WHEN three numbers are such that the first is to the third, as the difference of the first and second is to the difference of the second and third, they are said to be in HARMONICAL PROPORTION, and a series of numbers in continued harmonical proportion, constitutes a HARMONICAL PROGRESSION. The reciprocal of a number, is the quotient of 1 by the number. Thus is the reciprocal of 2; 4 is the reciprocal of is the reciprocal of 3, &c. The reciprocals of any equidifferent series form a harmonical proportion. PROBLEM I. Two numbers being given to find a third in harmonical proportion. * So called, because if a musical string be divided in harmonical proportion, the different parts will vibrate in unison. RULE. Consider the reciprocals of the numbers as two terms of The third term will be the reci an equidifferent series. procal of the number sought. Find a third harmonical proportional to 120 and 40. 5 The reciprocals are 120, and 40 or 30. The third term of the equidiffereat series is 1209 and its reciprocal 24 is the harmonical proportional sought. 1. The first two terms of a harmonical progression are 60 and 30. Required the ten succeeding terms. 2. The first two terms of a harmonical proportion are 348075 and 69615. Find the six succeeding terms. PROBLEM II. To insert any number of harmonical means between two numbers. RULE. Find as many arithmetical means between the reciprocals of the given numbers. These means will be the reci procals of the harmonical means. Insert 4 harmonical means between 20 and 120. 1 5 The reciprocals are 24 and 10 or 120 and 20. The four arithmetical means are 120, 120, 120 and 120, whose reciprocals are 24, 30, 40 and 60,-the desired harmonical means. 3. Insert 7 harmonical means between 630 and 5040. CHAPTER XX. ANNUITIES. ANY sum of money to be paid regularly, at stated periods, is called an ANNUITY. The payment may be stipulated for |