The first root figure in the following examples must be found by trial. 7. Five times the third power, plus three times the second power of a number, added to twice the number, make 9349968. What is the number? 8. A man being asked his age, replied : “ If 3 times the square of my age be added to 5 times my age, the sum will be 2668.” What was his age ? 9. 7 times the 4th power + 2 times the third power-13 times the second power of a certain number=2120912536. What is the number? 10. The 5th power of a certain number, diminished by 23 times the number, equals 14025514846. What is the number? 11. What is that number whose 5th power +5 times its 4th power-631 times its second power=42629137739? 12. What is that number 5 times whose 4th power+ 25 times its 3d power + 125 times its 2d power=1364615. 9375 ? 13. What is the number of miles from New York to Baltimore, if 13 times its 4th power+79 times its 3d power-93 times its 2d power,=17848292769 ? 14. The 6th power of a certain number exceeds its 2d power by 65944160560800. What is the number? 15. Extract the cube root of 3; the 5th root of 4 ; the 7th root of 5. 16. What is the value of x, if 5x5 + 3x4 +17x=16035686129037 ? When two equal quantities are connected by the sign of equality, as in the above example, the whole is called an EQUATION. Á QUADRATIC EQUATION, is one in which the highest power of the unknown quantity is the square. 17. Find the value of x in the quadratic equation 3x% +7x=780. 18. Find the root of the quadratic equation 2x2 +5x= 37125. 19. What is the value of x, if 9x2 +13x=58006? 20. What is the value of x in the quadratic equation, 4.32+x=159.3275 ? CHAPTER XVII. ence. ARITHMETICAL PROGRESSION-OR, EQUIDIF FERENT SERIES. A series of numbers, in which the successive terms in. crease or diminish uniformly by the same number, is called an ARITHMETICAL PROGRESSION,- PROGRESSION BY DIFFERENCE, or EQUIDIFFERENT SERIES. The difference between the successive terms, is called the common differ The first and last terms of the series are called the extremes; the other terms the means. Thus, in the ascending series, 2, 4, 6, 8, 10, 12, 14, 16, the extremes are 2 and 16, and the common difference is 2. In the descending series, 20, 16, 12, 8, 4, 0, the extremes are 20 and 0, and the common difference is 4. Any three of the five following things being given, the other two may be found : 1. The first term. PROBLEM I. One of the extremes, the common difference, and the number of terms being given, to find the other extreme and the sum of all the terms. What is the tenth term of an ascending series, the first term being 1 and the common difference 4 ? The second term will evidently be 1+4; the 3d, 1+2x4; the 4th, 1+3x4, and so on to the 10th, which is 1+9X4 or 37. What is the sum of the first ten terms of the above series? To obtain a rule for finding the sum, we will invert the whole series, and write it under itself. In this manner we shall evidently obtain twice the sum of the series. And we may moreover observe, that the sum of the extremes is equal to the sum of any two terms equally distant from the extremes, or to twice the middle term, when the number of terms is odd. 1, 5, 9, 13, 17, 21, 25, 29, 33, 37. 37, 33, 29, 25, 21, 17, 13, 9, 5, 1. 38, 38, 38, 38, 38, 38, 38, 38, 38, 38. We thus have 10 times 38, or the number of terms multiplied by the sum of the extremes, which is equal to twice the sum desired. Hence we derive the following RULE. To obtain the last term, multiply the common difference by the number of terms less one, and if the series is ascending, add the product to the first term,-if decreasing, subtract it from the first term. To obtain the sum of the terms, multiply the number of terms by the sum of the extremes, and divide the product by two.* 1. The first term of an equidifferent series is 5, the common difference 2, and the number of terms 11. Find the last term, and the sum of all the terms. 2. What is the sum of 45 terms of the natural series, 1, 2, 3, &c. ? 3. Find the 23d term, and the sum of 23 terms of the series 1, 4, 7, 10, &c. 4. How many times does a clock strike in 12 hours ? 5. The first term of a descending series is 30, the com. mon difference ž, and the number of terms 33. Required the last term, and the sum of the series. 6. If I travel 100 miles to-day, and 5 miles less on each succeeding day, how far shall I have gone at the end of the 17th day, and what will be the length of the last day's journey? PROBLEM II. The extremes and number of terms being given, to find the common difference. The first term of an equidifferent series is 27, the last term 3, and the number of terms 9. What is the common difference? The last term is found by Problem I. by subtracting 8 times * If the given term is the last term, invert the series, and work by the same rule. the common difference from the first term. Then the difference of the extremes 24, must be 8 times the common difference, which is therefore 3. RULE. Divide the difference of the extremes by the number of terms less one, and the quotient will be the common difference. This difference repeatedly added to the less, or subtracted from the greater term, will give the intermediate terms. The sum of the terms is found by the preceding problem. 7. One hundred and fifty eggs were laid in a row. The first was 2 yards, and the last 100 yards from a basket. How far were the eggs apart, and how far must a person travel to pick them up singly and lay them in the basket ? 8. A man had 15 children, whose ages were in Arithmetical Progression: the youngest was 3 years old, and the oldest 24. Required the common difference of their ages. 9. The extremes of an equidifferent series are 1 and 11, and the number of terms 25. What are the mean terms ? 10. Insert 12 arithmetical means between 20 and 59. [As there are 12 means, the number of terms must be 14.] 11. Insert 13 arithmetical means between 3 and 4ă. 12. A man commenced walking for exercise, daily in. creasing the distance by equal additions. The first day he walked 4 miles, and on the 15th day he walked 7} miles. What was the daily increase ? PROBLEM III. The extremes and common difference being given, to find the number of terms ? The extremes are 45 and 10, and the common difference is 5. What is the number of terms. By Problem I, the difference of the extremes is equal to the common difference multiplied by the number of terms less one. Then the difference of the extremes 35, divided by the common difference 5, gives 7, which must be equal to the number of terms less 1. Therefore the number of terms is 8. RULE. Divide the difference of the extremes by the common difference, and add 1 to the quotient. 13. What is the sum of the series 2, 4, 6, 8, &c., to 1000 ? 14. What is the sum of the descending series 100, 991, 99, 98ì, &c., the last term being 0 ? 15. The first term of a series is 11, the last term 2, and the common difference 1. What is the number of terms ? 16. The first term 23, the common difference š, and the last term 90, being given, required the number of terms. PROBLEM IV. The extremes and sum of the terms being given, to find the number of terms. RULE. Divide twice the sum of the terms by the sum of the extremes. 17. The sum of the terms is 221, the first term 4, and the last term 11. Required the number of terms and the common difference. PROBLEM V. One of the extremes, the number of terms, and the sum of the terms being given, to find the other extreme. RULE. Divide twice the sum of the terms by the number of the terms, and subtract the given extreme from the quotient. 18. The sum of the terms in an equidifferent series is 27, one of the extremes is 4, and the number of terms 8. Required the other extreme and the common difference. PROBLEM VI. The number of terms, common difference, and sum of the terms being given, to find the extremes., RULE. Divide the sum of the terms by the number of terms. Then multiply the common difference by the number of terms less one, and subtract half the product from the first quotient for one extreme. For the other extreme add half the product to the quotient. Problems IV., V., and VI., are all readily deduced from Problem I. |