6 EXAMPLE VI.-The extremes of a geometrical progression are 6 and 486, and the number of terms is 5 ; find the common ratio. The number of terms being 5, the last extreme is the product of the first extreme by the fourth power of the common ratio. So that the common ratio is the fourth root of the fraction ass, or of St. We obtain this root by extracting the square root of the square root of it: V=; V=}, the common ratio required. 264. To determine the means in a geometrical progression -- the extremes and the number of means being given : Find the common ratio ($ 263), remembering that the number of terms is greater by 2 than the number of means; and employ this common ratio, as multiplier, the necessary number of times ($ 260). EXAMPLE VII.—Insert 5 geometrical means between 3 and 12,288. The number of means being 5, the number of terms is (5+2=)7 ; so that 12,288 is the product of 3 by the sixth power of the common ratio. To find the common ratio, therefore, we divide 12,288 by 3, and extract the sixth root-or the cube root of the square root-of the quotient : 12,288+3= 4,096; V 4,096=64; 064=4, the common ratio. Multiplying 3 (the first extreme) by 4, we obtain the second term, 12; multiplying 12 by 4, we obtain the third term, 48; and so on. 'The required means are thus found to be 12, 48, 192, 768, ana 3,072. The progression in full is3 48 192 768 3,072 12,288. EXAMPLE VIII.-Insert one geometrical mean between 539 and 11. The number of terms being (1+2=)3, the last extreme is the product of the first extreme by the second power of the common ratio ; so that the common ratio is the square root of the fraction , or of 1. The square root of 15 being }, the required mean is 539x}=77. One geometrical mean, however, can be inserted more easily. 265. To insert ONE geometrical mean between two numbers : Multiply the numbers together, and extract the square root of the product. Thus, putting x for the geometrical mean between 539 and II, we have the proportion 539 : x :: X: 11. Consequently, z=539X11; I=V 539 XII=V 5929=77 T I2 20 NOTE._From this it follows that, of any three consecutive terms of a geometrical progression, the middle term is the square root of the product of the other two. 266. To find the sum of the terms of a geometrical progression—the extremes and the common ratio being given : Multiply the last extreme by the common ratio, and divide the difference between the product and the first extreme by the difference between the common ratio and unity. Thus, putting s for the sum of the terms of the progression 5 80 20,480; multiplying by the common ratio, 4; taking equals from equals; and dividing by 3 (the difference between the common ratio and 1), we have s=5+20+80+320+1,280+5,120+20,480 45= 20+80+320+1,280+5,120+20,480+81,920 _81.920-5. 35=81,920 - 5; s= =27,305. 3 Again: putting s for the sum of the terms of the progression 9 3 } 3 multiplying by the common ratio, 1 ; taking equals from equals; and dividing by $(the difference between the common ratio and 1), we have s=9+3+1+}++++al+ + 729 ; s=9*1=13381. 9-2187 NOTE 1.—When a geometrical progression is descending and interminate, the sum of the terms is obtained from the division of the first term by the difference between the common ratio and unity—the last term being regarded as o. Thus, writing the pure circulator •345 under the form 345 345 +&c., 345 345 +&c. 345 I 213 243 9 27 81 I of s=9–2,187 364 S = I +&c. + 1,000 1,000,000 1,000,000,000 XS= 999 345 X SE 1,000 1,000 345 S = 1,000 999 1,000 345 999 I 2 Note 2.-Every term (after the first) of the progression — 32 exceeds the sum of the preceding terms by 1. Thus, the second term (2) exceeds the first (1) by 1; the third term (4) exceeds the sum of the first two (1+2=3) by 1; the fourth term (8) exceeds the sum of the first three (i+2+4=7) by The fact is well known to professional gamblers that, after playing for and losing-first (say) £1, then £2, then £4, then £8, and so on, a person would, if he ultimately won a game, retrieve all his losses, and have £ i over. I; &c.* HARMONICAL PROGRESSION. 267. Four numbers are said to be in harmonical proportion when the first bears to the fourth the same ratio which the difference between the first and the second bears to the difference between the third and the fourth. Thus, 12, 6, 15, and 10 are in harmonical proportion, 12 bearing to 10 the same ratio which the difference between 12 and 6 bears to the difference between 15 and 10 : 12 : 10 :: 12 -6:15- 10. 268. Three numbers are said to be in harmonica) proportion when the first bears to the third the same ratio which the difference between the first and the second bears to the difference between the second and the third. Thus, 12, 8, and 6 are in harmonical proportion, 12 bearing to 6 the same ratio which the difference between 12 and 8 bears to the difference between 8 and 6: 12 : 6 :: 12–8:8–6. * The common ratio being 2, the nth term is (1 X 21~=)2n–), and the sum of n terms is 2n– 2- I; whilst the next term after the nth is (1 X 2" =)2", which exceeds 2" – I by 1. 269. A harmonical PROGRESSION—as already explained ($ 247)—is a series of numbers of which every three consecutive ones are in harmonical proportion. Thus, 120, 60, 40, 30, 24, and 20 form a harmonical progression, every three consecutive numbers being in harmonical proportion : 120 : 40 :: 120—60 : 60-40 30 : 20 :: 30–24 : 24-20 270. To insert a harmonical mean between two numbers : Divide the sum of the numbers into twice their product. Thus, taking any three consecutive terms of the harmonical progression 60 20, we find that the middle one is obtained as quotient when twice the product of the other two is divided by their sum: 2 X 120 X 40 60 2 x 60 x 30 120+40 60+30 Putting y for the harmonical mean between x and z, we have ·x: z:: 0;-Y: Y-z; xy-zz=x2-yz; xy+yz=xz+xz; (x+2) .XY=2x2; y: x-ta 271. By taking the reciprocals* of its terms, we convert a harmonical into an arithmetical progression, or an arithmetical into a harmonical progression—as the case may be. Thus, taking the reciprocals of the terms of the harmonical progression 60 40 30 24 20, we obtain the arithmetical progression TO 10 30 Zo 120 40 i 40 ; &c. 2xz a I 20 1 66 * Any two numbers whose product is unity are said to be the “reciprocals” of one another. Thus, j and are reciprocals of one another, jx being 1; 7 and are also reciprocals of one another, 7x7 being = I. So that, to find the reciprocal of a number, we divide unity by the number. The reciprocal of 5), for instance, or of 4, i , is (1;4=)it. 9 On the other hand, taking the reciprocals of the terms of the arithmetical progression 3 5 7 9 II, we obtain the harmonical progression f 1 ii Let x, y, and z be three consecutive terms of a harmonical progression, and let their reciprocals be x', y', and z', respectively: to prove that x', y', and z' are three consecutive terms of an arithmetical progression; in other words ($ 254), that it'tz' = 2y'. Here we have ($ 270) xy+yz=2xz; (dividing 1_!_2_2x!; i.e., z'tu'=zy'. 2y XYZ .xy? XYZ Y y Next, let x', y', and z' be three consecutive terms of an arith. metical progression, and let their reciprocals be x, y, and z, respectively: to prove that x, y, and z are three consecutive terms of a harmonical progression; in other words ($ 270), that Y= Here we have ($ 254) x' +'=2y'; (dividing by x'y':') x+z ' 2y' X-+ y'z' ' x'y' x'Z'' Y' ' ; i.e., y xz+xXy=2X*X2; (2+2) Xy=222; y=; zt by xyz) 2y + y2 I 272. To insert two or more harmonical means between a given pair of extremes: Between the reciprocals of the extremes insert as many arithmetical means as there are harmonical means to be determined ; and the reciprocals of the arithmetical means will be the harmonical means required. 5 This follows from $271. As an illustration, let it be re quired to insert three harmonical means between 36 and 4. The reciprocals of the given numbers are 3 and 1. Between these two fractions we insert ($ 253) three arithmetical means 1 371 the reciprocals of which are the harmonical means required 3.6 72 5? As a second illustration, let it be required to set down four 12 36 5 or 12 |