232. Twenty cases may arise in Geometrical Progression. In discussing these cases we shall preserve the same notation as in Arithmetical Progression, except that instead of d the common difference we shall use r the ratio. 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 7 required. The successive terms of the series are a, ar, ar2, ars, 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=arm-1. Hence, RULE. Multiply the first term by that power of the ratio whose index is one less than the number of terms. 4. Given a = — 7, ?= - 4, and n = 5. Given a =— }, r = = }, and n = 6. Given a = 5, r―― , and n = 3, to find l. Ans. 1: - 112. 5, to find 7. Ans. 1 —— 10, to find 7. 719. 7. Given a =— 8. Given a = }}, r = , and n = 8, to find l. 6, to find l. 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. Now S = a + ar+ar2 + ar3 +......+l (1) Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. 2. Given a = 7, 7=45927, and r 3, to find S. 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 = 7, and n = 5, to find S. 3. Given a = r= 1, and n = 7, to find S. 4. Given a = — 5, r=— - 4, and n = 5. Given a =— ț, r = 6, and n = 6. Given a = 4, to find S. Ans. S 255. 5, to find S. 236. In a geometrical series whose ratio is a proper fraction 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. 1. Find the sum of the series 1,,, &c. to infinity. 2. Find the sum of the series,,, &c. to infinity. 4. Find the sum of the series 6, 4, 23, &c. to infinity. Ans. 18. 5. Find the value of the decimal .4444, &c. to infinity. NOTE. This decimal can be written+ 180 + 1000, &c. 6. Find the value of .324324, &c. to infinity. Ans.. 7. Find the value of .32143214, &c. to infinity. CASE IV. 237. The extremes and number of terms given, to find the ratio. In this Case a, l, and n are given, and r is required. Divide the last term by the first, and extract that root of the quotient whose index is one less than the number of terms. 1. Given a =7,1=567, and n= 5, to find r. 3. Given a = — = 314, and n=4, to find ~. .5. NOTE. This rule enables us to insert any number of geometrical means between two numbers; for the number of terms is two greater than the number of means. Hence, if m = the number of means, m + 2 = n, or m + 1 = n 1; and r = found the ratio, the means are found by multiplying the first term by the ratio, by its square, its cube, &c. 4. Find three geometrical means between 2 and 512. 5. Find four geometrical means between 3 and 3072. Ans. 12, 48, 192, 768. 6. Find three geometrical means between 1 and 1. But ar is the second term of a series whose first term is a and ratio r; or the geometrical mean of the series a, ar, ar2. Hence, the geometrical mean between two quantities is the square root of their product. 7. Find the geometrical mean between 8 and 18. |