3. What is the sum of the numbers 1, 2, 3, 4, 5, &c., continued to 1000 terms? Ans. 500500. 4. What is the common difference of the terms in an Arithmetical progression whose first term is 10, last term 150, and number of terms 21? Ans. 7. 5. If the third term of an Arithmetical progression be 40, and the fifth term 70, what will the fourth term be? Ans. 55. 6. If the first term of an Arithmetical progression be 5, and the fifth term 30, what will the second, third, and fourth terms be? Find the common difference, (177), and thence the three intermediate terms. Ans. 11; 17; 23. 7. If the fourth term of an Arithmetical progression be 37, and the eighth term 60, what are the intermediate terms? Ans. 42; 48, 541. 8. What is the sum of 25 terms of an increasing Arithmetical progression in which the first term is, and the common difference of the terms also? (176). Ans. 162. 9. The first term of an increasing Arithmetical progression, is 1, and the number of terms 23. What must be the common difference, that the sum of all the terms may be 100? Let x represent the common difference; then 1+22x is the last term, (176); and 2+22x ×23 is the sum of the terms, (180). Hence an Equation may be formed from which the value of x will be found. Or we might substitute the numbers 1, 23, and 100, for a, n, and s in Formulas A and B, (181), and find the value of d, as one of the two unknown quantities. Ans. 10. If the first term of a decreasing Arithmetical progression is 100, and the number of terms 21, what must the common difference be, that the sum of the series may be 1260? Ans. 4. 11. A and B start together, and travel in the same direction; A goes 40 miles per day; B goes 20 miles the first day, and increases his rate of travel of a mile per day. How far will they be apart at the end of 40 days? Ans. 215 miles. 12. One Hundred stones being placed on the ground in a straight line, at the distance of 2 yards from each other; how far will a person travel who shall bring them, one by one, to a basket which is placed 2 yards from the first stone? Ans. 11 miles 840 yards. 13. Find the third term of an Harmonical progression whose first and second terms are 12 and 15 respectively. If a represent the third term, we shall have 12 x 15-12x-15. Ans. 20. Ans. 60. 14. What is the first term of an Harmonical progression whose second and third terms are 30 and 20 respectively? 15. What is the fourth term of an Harmonical proportion whose first, second, and third terms are 2, 3, and 8 respectively? Ans. 16. 16. If the first and third terms of an Harmonical progression be 25 and 40 respectively, what will the second term be? Ans. 3019 17. The first and fourth terms of an Harmonical progression, are 10 and 20 respectively. What are the two intermediate terms? This problem may be solved by finding two arithmetical means between and 20, and then taking the reciprocals of the terms thus found, (183). Ans. 12, and 15. 18. The fifth and eighth terms of an Harmonical progression are 20 and 40 respectively. What are the two intermediate terms? Ans. 24, and 30. 19. The first term of a Geometrical progression is 2, the ratio of the progression is 3, and the number of terms 4. What is the last term? and the sum of all the terms? Ans. 54, and 80. 20. The first term of a Geometrical progression is, the ratio of the progression is, and the number of terms 4. What is the last term? and the sum of all the terms? Ans. 4, and 29. 21. What is the sum of an infinite number of terms in the Geometrical progression whose first term is 100, and ratio ? Ans. 133. 22. What is the sum of an infinite number of terms in the Geometrical progression whose first term is 300, and ratio? Ans. 450. 23. If the first and third terms of a Geometrical progression are 8 and 72 respectively, what is the second term? The second term is equal to the square root of 8×72, (189). Or, considering the third as the last term of the progression, 72÷8-9 is the square of the ratio, (187); then 3 is the ratio of the progression; and the second term is now readily obtained. Ans. 24. 24. If the third and fifth terms of a Geometrical progression be 75 and 300 respectively, what will the fourth term be? Ans. 150 25. If the first and fourth terms of a Geometrical progression are 3 and 24 respectively, what are the two intermediate terms? Ans. 6 and 12. 26. If the seventh and tenth terms of a Geometrical progression are 6 and 750 respectively, what are the intermediate terms? Ans. 30 and 150. 27. What is the sum of an infinite number of terms in the series 1,,, &c., in which the ratio of the progression is evidently? Ans. 2. 28. If a body move forever at the rate of 2000 feet the first second. 1000 the second, 500 the third, and so on, what is the utmost distance it can reach ? Ans. 4000. 29. If 10 yards of cloth be sold at the rate of $1 for the first yard. $2 for the second, $4 for the third, and so on, what would be the price of the last yard? and what would the whole amount to? Ans. $512, and $1023. 30. If 13 acres of land were purchased at the rate of $2 for the first acre, $6 for the second, $18 for the third, and so on, what would the last acre amount to? Ans. $1062882. 31. Allowing the interest of a sum of money to be $500 the first year, $400 the second, $320 the third, and so on, forever, what would be the whole amount of interest? Ans. $2500. 32. Two bodies move at the same time, from the same point, in opposite directions. One goes 2 miles the first hour, 4 the second, 6 the third, and so on; the other goes 2 miles the first hour, 4 the second, 8 the third, &c.; how far will they be apart at the end of 12 hours? Ans. 8346 miles. 33. A and B set out at the same time to meet each other. A travels 3, 4, 5, &c. miles on successive days, and B 3, 4, 62, &c. miles on successive days. They meet in 10 days; what is the distance between the two places from which they traveled? Ans. 414 miles. 121 CHAPTER VIII. PERMUTATIONS AND COMBINATIONS.-INVOLUTION.-BINOMIAL THEOREM.-EVOLUTION. PERMUTATIONS. (193.) PERMUTATIONS are the different orders of succession in which a given number of things may be taken-either the whole number together, or the whole number taken two and two, or three and three, &c. Thus the different Permutations of the three letters a, b, and c, when all are taken together, are abc, acb, bac, cab, bca, cba. And the different Permutations of the same letters when taken two and two, are ab, ba, ac, ca, bc, cb. Number of Permutations. (194.) If n represent a given number of things, the number of mutations that can be formed of them, will be equal to n(n−1)(n—2)(n—3)(n—4), and so on, per until the number of factors multiplied together is equal to the number of things taken in each permutation. To demonstrate this proposition, suppose that we have n letters, a, b, c, d, &c., to be subjected to Permutations. If we reserve one of the letters, as a, there will remain n-1 letters; and by writing a before each of the remaining letters, we have n-1 permutations of n letters, taken two and two, in which a stands first. In like manner we should find n-1 permutations of n letters, taken two and two, in which ỏ stands first; and so for each of the n letters. Hence we shall have n(n-1) permutations of n letters taken two and two. Suppose now that the n letters are to be taken three and three. By reserving a, and proceeding with n-1 letters as before, we should find (n-1)(n-2) permutations of n-1 letters taken two and two; and by writing a before each of these permutations, we have (n-1)(n-2) permutations of n letters, taken three and three, in which a stands first. We should find the same number of permutations of n letters, taken three and three, in which 6 stands first; and so for each of the n letters. Hence we shall have n(n−1)(n—2) permutations of n letters taken three and three. Suppose now that the n letters are to be taken four and four. By reserving a, and pursuing the operation in the same manner as before, we should find n(n−1)(n—2)(n—3) permutatims of n letters taken four and four. Thus the demonstration proceeds; the number of factors multiplied together being found always equal to the number of letters taken in each permutation. As an Example of the application of the principle above demon strated, suppose it were required to determine the number of Permutations, or different orders of succession, that could be formed in a class composed of six pupils, by taking the whole number in each permutation. Since the six pupils are to be taken in each Permutation, the number of factors to be employed is six; hence the number of permutations is 6×5×4×3×2×1=720. If the whole six were to be subjected to Permutations by taking five at a time, the number of permutations would be 6x5x4x3x2=720. If the whole six were to be subjected to Permutations by taking four at a time, the number of permutations would be 6x5x4x3=360. It will be observed above, that the number of Permutations will be the same, whether the whole number of things, or one less than the whole number, be taken in each permutation. |