S= 361. To find the sum of a decreasing geometrical series, when the number of terms is infinite. By changing signs in the numerator and denominator, equation (B) may be written a(1—p) 1.-p Now suppose r less than unity; then the larger n is, the smaller will not be; and by making n large enough, may be made less than any assignable quantity, or zero. Hence, if the number of terms is infinite, you may be neglected in comparison with unity; and we shall have as the formula for an infinite series, S . (C) 1362. To insert a given number of geometrical means between two given quantities. Let n' denote the number of means to be inserted; then the whole series will consist of n'+2 terms. Hence, putting n= = n'+2 in equation (A), we have 1= ar+1. Whence we obtain, W+1 11 r = Having found the ratio, the required means may be obtained. 363. Since the terms of a geometrical series, taken consecutively, have the same ratio one to another, it follows that they are in continued proportion; (327). Hence, 1st. When three terms are in geometrical progression, the product of the extremes is equal to the square of the mean. 2d. When four terms are in geometrical progression, the product of the means is equal to the product of the extremes. APPLICATION OF THE FORMULAS. 364. The two primitive equations, 'a[mm_1) l= ar 1 S = r-1 contain the five quantities, a, r, I, n, S, any three of which being given, the other two may be found; for, by substitution of the given values, the result will always be two equations involving but two unknown quantities. In this general problem there will be ten cases, as in the corresponding problem of Arithmetical Progression. We can not, however, obtain a solution of all the cases, by simple or quadratic equations. 1st. The quantity n enters the two equations only as an exponent, and its value can not be obtained by the common methods of reducing an equation. The process involves the principle of logarithms, and will be presented in its proper place. 2d. The quantity r is affected by an exponent in both equations; and its value must be obtained by extracting the (n-1)th or the nth root of a quantity. When n is not large, r can readily be found by inspection or trial. 3d. The values of a, l, and S will be found by means of simple equations, as in Arithmetical Progression. 1. The first term of a geometrical progression is 3, and the ratio 2 ; find the 12th term, and the sum of the series. We have given a = 3, r= 2, Whence by formulas (4) and (B), 1 = 3X2" = 3X2048 = 6144 Ans. n = 12. S= n = 2. The sum of a geometrical progression is 1820, the number of terms 6, and the ratio 3 ; find the first term, and the last term. We have given, S= 1820, = 6, r = 3. By formula (B), a(3-1) = 364a; 3-1 a = 5, first term. Then by formula (A), l= 5x3 = 1215, last term. 1820 = 3. It is required to find 3 geometrical means between 6 and 486. By formula (D), we have V 186 = =Vg1 = 3. Therefore, the series is 6, 18, 54, 162, 486, Ans. 4. Find the sum of the series 6, 2, į, ž.... to infinity. We have given, a = 6, r=1; hence, by formula (C), 6 S= Ans. -9, 3 + etc. 5. Find the exact value of the decimal .454545.... to infinity. 45 10000 1000000 In all such cases, the repetend, taken with its local value, will be the first term of a geometrical series, of which the ratio will be 10 or some power of 10. In the present example we have 45 1 hence, 100' 100 a = 1. Find the sum of 9 terms of the series 1, 2, 4, 8, Ans. 511. 2. Find the 8th term of the progression 2, 6, 18, 54, Ans. 4374. 3 Find the sum of 10 terms of the series 1, 3, $, 27, Ans. 474075. 89047 4. Required two geometrical means between 24 and 192. Ans. 48, 96. 5. Required 7 geometrical means between 3 and 768. Ans. 6, 12, 24, 48, 96, 192. 6. Find the value of 1+&+48 + x1 + o... to infinity. Ans. 4. 7. Find the value of +1+1+ 2+ .... to infinity. Ans. 44. 8. Find the value of 5+1 + $ + +.... to infinity. Ans. 74 9. Find the value of the decimal .323232.... to infinity. Ans. 33 10. Find the value of the decimal .212121.... to infinity. Ans. 35 11. Find the value of 1-1+1-istas - .... to infinity. Ans. s 12. Find the value of } - 2's tiis – 616 t ....to infinity. Ans. Ans. a'x 15. The sum of a geometrical series is 1785, the ratio 2, and the number of terms 8; find the first term. Ans. 7. 16. The sum of a geometrical series is 7812, the ratio 5, and the number of terms 6; find the last term. Ans. 6250. 17. The first term of a geometrical series is 5, the last term 1215, and the number of terms 6. What is the ratio ? Ans. 3. 18. A man purchased a house with ten doors, giving $1 for the first door, $2 for the second, $4 for the third, and so on. What did the house cost him ? Ans. $1023. PROBLEMS IN GEOMETRICAL PROGRESSION TO WHICH THE FORMULAS DO NOT IMMEDIATELY APPLY. 365. The terms of a geometrical progression are represented in a general manner as follows: X, xy, xy', wy",.... In the solution of problems, however, the following notation is generally preferable : 1st. When the number of terms is odd, the series may be reprcsented thus : y; xy, 2d. When the number of terms is even, the series may be expressed thus : yo X, Y, ; C V xy, squares 364. yo y X, Y, y y We e may also represent three terms as follows: X, y. 1. The sum of three numbers in geometrical progression is 26, and the sum of their What are the numbers ? (1) (2) Transposing V xy in (1), squaring and reducing, x*+xy+y=a'—2aV zy. (3) From (2) and (3), a-2aV xy = b; a2b whence, 2a From (1) and (2) x = 2, and y Hence, the numbers are, 2, 6, 18, Ans. . = 6. 1.8. |