The first term is the product of the last by a power of the common ratio, the index of which power is less by 1 than the number of terms. We therefore divide 234,375 by 3, and find -in either of the two ways shown below-what power of 5 the quotient (78,125) is. The index of the power being 7, the number of terms is (7+1)8:-* 263. The extremes of a geometrical progression and the number of terms being given, to find the common ratio: Divide the last extreme by the first, and evolve from the quotient the root indicated by what remains when the number of terms is diminished by I. * We obtain this result almost-if not quite-as readily by actually calculating the intermediate terms, in the way shown in the margin; so that § 262 is practically useful only when the number of terms is large, and a table of Logarithms can be employed. 1875 9375 46875 234375 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 or of We obtain this root by extracting the square root of the square root of : ; √, 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; 4,096-64; 64-4, 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 is— 3 12 48 192 768 3,072 12,288. EXAMPLE VIII.-Insert one geometrical mean between 539 and II. 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 339, or of. The square root of 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 : II. Consequently, x2=539X11; x=539 x 11=5929-77. T 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 20 80 320 1,280 5,120 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 20+80+320+1,280+5,120+20,480+81,920 4s= Again: puttings for the sum of the terms of the progression 9 3 A 215 multiplying by the common ratio,; taking equals from equals; and dividing by (the difference between the common ratio and 1), we have s=9+3+1+++tatelatron } of s= 27 81 729 NOTE I. 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 + 345 + &c., and putting s for the sum of all the fractions, we have— NOTE 2.-Every term (after the first) of the progression I 16 32 &c. 2 4 8 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 I; the fourth term (8) exceeds the sum of the first three (1+2+4=7) by I; &c.* 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 £1 over. 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 harmonical 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: 126: 12-8:8-6. *The common ratio being 2, the nth term is (1 x 2"'—)2”—1, and the sum of n terms is ( 2"-1; whilst the next term 2-I after the nth is (1 × 2" =)2", which exceeds 2" -1 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 60: 30: 60-40 40-30 270. 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 120 60 40 30 24 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 60 x 30. &c. 60: = 2 X 120 X 40 60+30 Putting y for the harmonical mean between x and z, we have x: z :: x-y: y−z; xy-xz=xz-yz; xy+yz=xz+xz; (x+2) 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 120 60 40 30 24 we obtain the arithmetical progression * 20, 120 80 18 30 2 20. = Any two numbers whose product is unity are said to be the "reciprocals" of one another. Thus, and are reciprocals of one another, xbeing: 1; 7 and are also reciprocals of one another, 7× So that, to find the reciprocal of a number, we divide unity by the number. The reciprocal of 5, for instance, or of , is (1÷V=). being = I. |