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 +10 + 10‰o, &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. 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 m+117 means, m + 2 = n n, or m+1 = 1; and r = V. Having α 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 T. Ans.,,. NOTE. When m= = 1, the formula becomes Multiplying by a, r= ara a 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. Ans. 12. 8. Find the geometrical mean between 4 and 343. 238. From the formulas established in Arts. 233 and 234, can be derived formulas for all the Cases in Geometrical Progression. From (1) we can obtain the value of any one of the four terms, l, a, n, or r, when the other three are given; from (2), the value of S, l, r, or a, when the other three are given. Formulas for the remaining twelve Cases which may arise are derived by combining the formulas (1) and (2) so as to eliminate that one of the two unknown terms whose value is not sought. 1. Find the formula for the value of S, when I, n, and r are given. S rn-1 r-1 l (TM — 1) (r — 1) pn − 1 NOTE. The four formulas for the value of n cannot be derived or used without a knowledge of logarithms; and four others, when n exceeds 2, cannot be reduced without a knowledge of equations that cannot be reduced by any rules given in this book. 239. To find any one of the five elements when three others are given. RULE. Substitute in that one of the formulas (1) or (2) that contains the four elements, viz. the three given and the one required, the given values, and reduce the resulting equation. If neither formula contains the four elements, derive a formula that will contain them, then substitute and reduce the resulting equation; or substitute the given values before deriving the formula, then eliminate the superfluous element and reduce the resulting equation. Substituting the given values of r, n, and S in (1) and (2), we obtain (3) and (4); finding the value of a, the superfluous element, from (3) and (4), and putting these values equal to each other, we form (7), an equation containing but one unknown quantity. Reducing (7) we obtain (9), or l = = 486. 2. Given a = 4, r = 5, and S15624, to find 7. 3. Given a = 2, n5, and 7 512, to find S. 4. Find the formula for the value of a, when r, n, and S are given. (r-1) S Ans. a= - 1 5. A gentleman purchased a house, agreeing to pay one dollar if there was but one window, two dollars if there were two windows, four if there were three, and so on, doubling the price for every window. There were 14 windows. How much must he pay? Ans. $8192. 6. A man found that a grain of wheat that he had sown had produced 10 grains. Now if he sows the 10 grains the next year, and continues each year to sow all that is produced, and it increases each year in tenfold ratio, how many grains will there be in the seventh harvest, and how many in all? In 7th harvest, 10000000 grains. In all, 11111111 grains. Ans. PROBLEMS TO WHICH THE FORMULAS DO NOT DIRECTLY APPLY. 240. In solving Problems in Geometrical Progression, if we let x = the first term and y = the ratio, the series will be x, xy, xy2, xy3, &c. It will often be found more convenient to represent the series in one of the following methods: — 1st. When the number of terms is odd, Which method is most convenient in any case will depend upon the conditions that are given in the problem. 1. There are three numbers in geometrical progression, the greatest of which exceeds the least by 32; and the difference of the squares of the greatest and least is to the sum of the squares of the three as 80:91. What are the numbers? SOLUTION. Let x, xy, and x y represent the series. Then = xy-x=32 (1) x2y1—x2: x2+x3y2+x2y — 80:91 x=4 (7) (2) (3) 91 9180+80y+80y (4) y-1:1+y+y=80:91 |