Let x the mean term and y the common difference; then the series will be x y, x, and x+y. 2. The sum of four numbers in arithmetical progression is 44, and the sum of the cubes of the two means is 2926. Ans. 5, 9, 13, 17. 3. Find seven numbers in arithmetical progression such that the sum of the first and fifth shall be 10, and the difference of the squares of the second and fourth 40. 4. There are four numbers in arithmetical progression; the product of the first and third is 20; and the product of the second and fourth 84. What are the numbers? 5. The sum of four numbers is 32; and their product 3465. Ans. 2, 6, 10, 14. in arithmetical progression What are the numbers? Ans. 5, 7, 9, 11. 6. The sum of the squares of the extremes of four numbers in arithmetical progression is 461; and the sum of the squares of the means 425. What are the numbers? Ans. 10, 13, 16, 19. 7. A certain number consists of three figures which are in arithmetical progression; if the number is divided by the sum of its figures, the quotient will be 15; and if 396 is added to the number, the order of the figures will be inverted. What is the number? Ans. 135. 8. Find four numbers in arithmetical progression such that the sum of the squares of the first and third shall be 104, and of the second and fourth 232. 9. Find four numbers in arithmetical progression such that the sum of the squares of the first and second shall be 29, and of the third and fourth 185. SECTION XXIII. GEOMETRICAL PROGRESSION. 229. A GEOMETRICAL PROGRESSION is a series in which each term, except the first, is derived by multiplying the preceding term by a constant quantity called the ratio. 230. If the first term is positive, when the ratio is a positive integral quantity, the series is called an ascending series, and when the ratio is a positive proper fraction, a descending series; but if the first term is negative, the series is ascending when the ratio is a positive proper fraction, and descending when the ratio is a positive integral quantity. Thus, 18, 6, 2, are ascending series; 32, -64, &c. are descending series. If the ratio is negative, the terms of the progression are alternately positive and negative. Thus, if the ratio is - 2 and the first term 3, the series will be 3, 6, 12, 24, 48, &c.; - - 3, 12, + 24, 48, &c. The positive terms of these two series constitute an ascending progression whose ratio is the square of the given ratio; and the negative terms a descending progression having the same ratio. 231. In Geometrical Progression there are five elements, any three of which being given, the other two can be found. These elements are the same as in Arithmetical Progression, except that in place of the common difference we have the ratio. 232. Twenty cases may arise in Geometrical Progression. In discussing these cases we shall preserve the notation as in Arithmetical Progression, except that instead of d the common difference we shall use r the ratio. same notation = CASE I. 233. The first term, ratio, and number of terms given, to find the last term. In this Case a, r, and n are given, and I required. The successive terms of the series are a, ar, ar2, ar3, art, &c. That is, each term is the product of the first term and that power of the ratio which is one less than the number of that term counting from the left; therefore the last or nth term in the series is or arn-1 l=ar-1. Hence, RULE. Multiply the first term by that power of the ratio whose index is one less than the number of terms. 1. Given a = 7, r=3, and n = 5, to find l. 3. Given a = 64, r = , and n = 10, to find 7. 4. Given a = - 7, r = - 4, and n = 3, to find l. Ans. 1: — 112. CASE II. 234. The extremes, and the ratio given, to find the sum of the series. In this Case a, l, and r are given, and S is required. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. 1. Given a = 2, 720000, and r 10, to find S. 4. Given a =— 1715, 7=, and r—— 7, to find S. CASE III. Ans. S60 343. 235. The first term, ratio, and number of terms given, to find the sum of the series. In this Case a, r, and n are given, and S required. The last term can be found by Case I., and then the sum of the series by Case II. Or better, since Substituting this value of Ir in the formula in Case II. we have RULE. From the ratio raised to a power whose index is equal to the number of terms subtract one, divide the remainder by the ratio less one, and multiply the quotient by the first term. 1. Given a = 4, r = S= r 7, and n = 75 5, to find S. Xa = 7x4=11204, Ans. 2. Given a = +, r = = 5, and n = 6, to find S. Ans. S 558. 236. In a geometrical series whose ratio is a proper frac tion the greater the number of terms, the less, numerically, the last term. If the number of terms is infinite, the last term must be infinitesimal; and in finding the sum of such a series the last term may be considered as nothing. Therefore, when the number of terms is infinite, the formula Hence, to find the sum of a geometrical series whose ratio is a proper fraction and number of terms infinite, RULE. Divide the first term by one minus the ratio. |