QUESTIONS RELATING TO ARITHMETICAL PROGRESSIONS, RESOLVED BY EQUATIONS. EXAMPLE 1. The first term of an arithmetical progression is 8; the last, 43; and the sum of the terms, 204. What is the common difference? The first term, the last, the common difference, the number of terms, and their sum, being represented, as before, by the letters a, l, d, n, and s, we have, from Pr. III., Then, by Pr. II., d÷35÷7=5, Ans. EXAMPLE 2. The sum of five numbers in arithmetical progression is 90; and the sum of their squares, 1710. What are the numbers? Let x represent the middle term, and y the common difference. Then the five numbers will be represented by x-2y, x−y, x, x+y, and x + 2y; and their squares will be x-4xy+4y2, x2-2xy + y2, x2, x2 + 2xy + y2, and x2+4xy+4y2. The sum of the five numbers is 5; and the sum of their squares, 5x2 + 10y2. ..5x90, x=18, and 5x2=1620. From this we 18, 21, and 24. or 1620 + 10y2 = 1710. obtain y=3; and the five terms, 12, 15, EXAMPLE 3. There are four numbers in arithmetical progression, such that the product of the second and third is 2250; and that of the first and fourth 2200. What are the numbers? Let the numbers be denoted by x-3y, x-y, x+y, and x+3y, the common difference being 2y. Hence we find y2 = 64, 2y=5, and x = 471. Therefore the four numbers are 40, 45, 50, and 55. NOTE. It will be seen, from the preceding examples, that there are various ways of expressing the terms, suit able to different questions. The best methods are soon discovered after a little experience. The object of the peculiar modes adopted in Examples 2 and 3, is, it will be perceived, to get rid of the product xy, in all the equations. Both questions, however, might have been resolved almost equally easily by expressing the terms, x, x + y, x+2y, &c. In the preceding examples, and in the following exer cises, the progressions are understood to be ascending, unless otherwise expressed; but the processes are almost exactly the same in the case of descending progressions. EXERCISE A. The first term is 10; and the last, 38. The sum of the terms is 120. What is the number of terms? 1. The sum of 20 terms is 200; and the last term, 29. What is the first? B. The first term is of all the terms is 517. Ans. -9. 77; and the last, 17. The sum What is the second term? C. Of three numbers in arithmetical progression, the difference of the squares of the first and second is 20; and of the second and third, 28. What is the progression ? D. The sum of the four terms of an arithmetical progression is 80, and the sum of their squares 1620. What are the terms? E. Of five numbers in arithmetical progression, the product of the first and fifth is 48; and that of the second and fourth, 60. What is the progression? 2. The common difference is 8; the number of terms, 9; and the product of the first and last, 2112. What is the first term? Ans. 24. F. In an arithmetical progression of four terms, the sum of the squares of the two mean terms is 1409; and that of the two extremes is 1445. Compute the numbers. 3. The product of four numbers in arithmetical progression is 98560, their common difference being 6. What is the first term? Ans. 10. G. Of six numbers in arithmetical progression, the common difference is 4, and the ratio of the first to the last is that of 2 to 3. What are the numbers? CHAPTER XXIV. (22). GEOMETRICAL PROGRESSIONS. DEFINITIONS. If a series of quantities increase, from continual multiplication by the same quantity, or decrease, from continual division by the same quantity, the series is called a Geometrical Progression. as 2, 6, 18, 54, 162, 486; or 64, 32, 16, 8, 4, 2. Even when the series decreases it may be considered to do so by multiplication, the multiplier, in that case, being a fraction. Consequently every geometrical progression may be expressed thus, a, ar, ar2, ar3, Regarding all geometrical progressions in the latter point of view, viz. as formed by successive multiplications, the constant multiplier, r, is called the common Ratio of the terms. As usual n expresses the number of terms. EXERCISE A. What are the values of r and n in each of the two numerical progressions given above? B. Form a geometrical progression of five terms, having its first term 7a, and the common ratio of the terms 2. C. Form a geometrical progression of four terms, having its first term 4x, and the common ratio 5y. D. The first term of a geometrical progression is y, the second y÷ 3, and the last z. What is the last but one? E. If the second and third terms are 2 and 2, what are the first and fourth? y NOTE. The successive terms of a geometrical progression are continual proportionals: that is, the first is to the second as the second to the third, as the third to the fourth, and so on: for a ar ar: ar2 :: ar2 : ar3, and so on. Consequently any term, standing between two others, is a mean proportional between them. EXERCISE F. Fill up the blanks in the following geo. metrical progressions :— 3, 12, 9a, - 81a. THEOREM. In any Geometrical Progression the Product of the two extreme Terms is equal to the Product of any two Terms equidistant from them; and also equal to the Square of the middle Term when the Number of Terms is odd. DEM. If the first term be a and the last 1, the progression may be expressed thus: there is an odd number of terms, the middle term may be expressed as either armor; and again, arTM × Or, if the progression be expressed thus,- al. ar"-3, ar-2, ar", then a, ar, ar2, a × ar”-1= a2,”—1, ar × ar”—2±a2μ”—1, ar2 × ar2¬3± a2μ»–1, and so on, all the products being alike. If there is an odd n-1 number of terms, the middle term will be ar 2, the square of which is a2-1, as before. NOTE. Any part of a geometrical progression is, of course, a geometrical progression. EXERCISE 1. In an increasing geometrical progression, consisting of seven terms, the first is 13, and the middle one 104. What is the last? Ans. 832. 2. If the first term is 7a, and the fifth 567ab4, what are the intermediate terms? Ans. 21a2b, 63a8b2, and 189a1b3. A. In a geometrical progression the first term is fifth, and the sixth 14. What is the tenth? the B. The first term of a geometrical progression is 3x, and the fifth term 4x5 3' What is the ninth? PROBLEM I. Having given the first Term of a Geometrical Progression, and the common Ratio, to find any other Term (say the nth). RULE. Raise the common ratio to a power whose ex ponent is a unit less than the number of the term required. viz. n—1, and multiply that power into the first term. DEM. The rule follows immediately from the definition of a geometrical progression. For, by that definition, if the first term be a, the common ratio r, and the number of terms n, the progression will be á, ar, ar2, ar3, in which any term is equal to the product of a into that power of r whose index is 1 less than the number of the term. EXAMPLE. The first term of a geometrical progression being 50, and the common ratio 7, what is the fourth term? 78 x 50343 x 50=17150. Ans. EXERCISE 1. The first term of a geometrical progression is 13; and the common ratio, 2. What is the tenth term? A. The first term of a geometrical progression being 0016, and the common ratio 5a, what is the tenth term? et 16' B. The first term being and the common ratio what is the ninth term? 2 3. What is the amount of £488, for four years, at 5 per cent. per annum, compound interest? Ans. £593-16705. C. What is the amount of £200, for 5 years, at 4 per cent. per annum, compound interest? PROBLEM II. Having given the first Term of a Geometrical Progression, the last Term, and the common Ratio, to find the Sum of all the Terms. RULE. Divide the difference of the first term and the last by the difference of the common ratio and unity. To the quotient add the last term. DEM. Expressing the progression in the general form shown in the Definitions, and putting s for the sum of the |