3. A person had seven sons, who had each saved a certain number of shillings, which were in arithmetical progression: the number saved by the second was 8, and the number saved by the oldest was 18. Required the savings of each. Ans. 6, 8, 10, 12, 14, 16, and 18. 4. There is a number consisting of three digits, which are in arithmetical progression, which being divided by the sum of its digits, the quotient is 283; but if 198 be added to it, the digits will be inverted. Required the number. Ans. 345. 5. The sum of four numbers, in arithmetical progression, is 56, and the sum of their squares = 864. Required the numbers. Ans. 8, 12, 16, and 20. 6. Find four numbers in arithmetical progression, the sum of the squares of the means being 400, and the sum of the squares of the extremes 464. Ans. 8, 12, 16, and 20. 7. After A, who travelled four miles an hour, had been on his journey 24 hours, B set out to overtake him by going 4 miles the first hour, 43 the second, 5 the third, &c. How many hours was B in overtaking A? Ans. 8. 8. The sum of the first and second of three numbers in geometrical progression is 12, and the sum of the second and third is 36. Required the numbers. Ans. 3, 9, and 27. 9. It is required to find three numbers in geometrical progression, whose sum is 7, and the sum of their squares 21. Ans. 4, 2, and 1. 10. Required three numbers in geometrical progression, whose product is 27, and the sum of their squares 91. Ans. 1, 3, and 9. 11. A person has four sons, whose ages are in arithmetical progression, the product of the youngest son's age multiplied by that of the oldest is 40, and the quotient of the third son's age divided by the second is 14. Required each son's age. Ans. 4, 6, 8, and 10 years. 12. A stationer sold seven books, their prices in shillings being in decreasing arithmetical progression; the price of the second was 13s., and the number of shillings that the first cost, was to the number the seventh cost, as 5 to 1. Required the price of each. Ans. 15s., 13s., 11s., 9s., 7s., 5s., and 3s. 13. Required four numbers in geometrical progres sion, the sum of whose means is 30, and the sum of the extremes 45. Ans. 5, 10, 20, and 40. par 14. A person bought four reams of paper, the ticular prices of which (in shillings) were in geome trical proportion: the number of shillings the first cost, multiplied by the number the second cost, makes £1. 12s., and the product of the third and fourth, multiplied in shillings, is £25. 12s. Required the price of each. Ans. The first 4s., second 8s., third 16s., and fourth 32s. 15. It is required to distribute £4. 10s. among four persons, A, B, C, and D, so that their respective shares, in shillings, may be in geometrical progression; the sum received by A and C was 30s. Required the sum received by each. Ans. A received 6s., B 12s., C 24s., and D 48s. 16. Required four numbers in geometrical progression such, that if the first and second be multiplied by 2, and 6 added to the third, they will form a series of numbers in arithmetical progression. Ans. 3, 6, 12, and 24. 17. A person bought four remnants of cloth, whose respective lengths in yards were in arithmetical progression, and found that if the first had contained 2 yards, the second 3 yards, the third 9 yards, and the fourth 25 yards more than they did, their respective lengths would have been in geometrical progression. How many yards did he buy? Ans. 36 yards. 18. What five numbers are in geometrical progression, the sum of the first and third being 15, and the sum of the third and fifth 60? Ans. 3, 6, 12, 24, and 48. 19. Two travellers, A and B, set out together from the same place; A goes eight miles the first day, twelve the second, &c., B goes one mile the first day, four the second, nine the third, &c. according to the square of the number of days he travelled. In what time will B overtake A? Ans. 7 days. 20. Two persons, A and B, at the distance of 462 miles, set out at the same time to meet each other: A goes one mile the first day, two the second, three the third, &c., and B travels each day a number of miles which is equal to the cube of the number of miles that A travels on the same day. In what time will they meet? Ans. 6 days. 147 SOLUTION OF CUBIC AND OTHER EQUATIONS BY APPROXIMATION. DEFINITIONS. 1. A Cubic Equation is one that contains the third power of the unknown quantity, as x3+ax2+ bx=c. 2. A Biquadratic Equation is one that contains the fourth power of the unknown quantity, as x1+ax3+ bx2+cx=d. 3. An Equation of the fifth power, is one that contains the fifth power of the unknown quantity, &c. Many ingenious rules, though some of them very tedious, have been given for the solution of cubic and other equations of higher powers, but they may be more readily solved by approximation, sometimes called "trial and error." RULE. Substitute for x in the given equation, two numbers as near its true value as possible, observing that one of them be greater and the other less than its true root or value, and mark the separate results; then, as the difference between these results is to the difference between the numbers substituted, so is the difference between the true result and either of the former to the respective correction of each: this correction being added to the number when too small, or subtracted from it when too large, will give the root or value of x nearly. The number thus found, with any other that may be supposed to approach still nearer to the true root or value of æ, may be assumed for another operation, which may be repeated till the root shall be determined to any degree of exactness that may be required. EXAMPLES. 1. Given x3+x2+x=100, to find the value of x. Here, by a few trials, x is found to be more than 4, and less than 5; therefore, if these two numbers be substituted in the given equation, and the results calculated as follows: as 71: 1 :: 16: .2253+ therefore, x=4+.2253-4.2253, which approximates nearly to the true value. If now 4.2 and 4.3 be taken as the assumed numbers, and the operation repeated, 6.369.1 :: 2.297: .036; therefore, x=4.3-.036-4.264, nearly. |