CHAPTER III PROGRESSION: ARITHMETIC, GEOMETRIC, AND HARMONICAL ARITHMETIC PROGRESSION 519. Definition.-An arithmetic progression is a series of numbers, which is so constituted that each number is equal to the preceding increased, or decreased, by a constant number, which is called the common difference. The various numbers of the series are the terms of the progression. The progression is increasing, when the terms continually inIn this case the common difference is positive. The increasing progression crease. 3, 7, 11, 15, 19, 23, etc., has the common difference +4. On the contrary, the progression is decreasing, when the terms continually diminish. the common difference is negative. In this case, The decreasing progression Thus the common difference is regarded as positive when the progression is increasing, and as negative in the contrary case; for this reason every term of the progression is the algebraic sum of the term which precedes it and the common difference. E. g., in the first example above, and in the second example 19 = 15+4, 611(5), and 9 −4+ (−5). 520. If a series of numbers a, b, c, d, is in an arithmetic progression, this fact may be indicated by the abbreviation A. P. 521. PROBLEM I.-Given the first term and the common difference of an arithmetic progression; calculate the value of the nth term. Let a be the first term and d the common difference of the A. P. By definition, the 2nd term is equal to the 1st term increased by d, or a + d the 3rd and so on; so that any term in the progression is equal to the first term plus the product of the common difference by the number of Hence if is the nth term of an the term preceding the required term. A. P., that is the term which has n determined by the formula (i) 1 terms preceding it, I will be 522. Application.-A body falls at Cincinnati 4.902 meters in the first second, and in any succeeding second 9.804 meters more than in the preceding; if the body falls for 6 seconds find the distance through which it falls during the sixth second. According to formula (i) the distance will be 4.902m+ (6 1)9.804m or 53.922m. What is the 12th term of the A. P. Here 1 1 1 6 523. In an increasing A. P., the terms increase without limit; that is, n can be taken so large that the nth term is greater than any given quantity A. For this purpose it is sufficient that In case of an increasing A. P., if the nth term is greater than A, then it follows that every term which follows the nth term is also greater than A. EXAMPLE. In the A. P. 5, 15, 25, 35, . . . etc., what values must n have in order that the nth term may be greater than 5000? According to formula (ii) i. e. n> 1+ - 5000 5 n 1+ 499.5 or 500.5; it is sufficient therefore if n is equal to or greater than 501. 524. THEOREM I.—In an A. P. composed of a limited number of terms the sum of two terms equally distant from the extreme terms is constant, i. e., independent of d. Consider the A. P. (a) a, b, c, p, g, h, k, l, composed of n terms in which the common difference is d. If the order of the terms of the A. P. (a) is reversed, a new A. formed, namely P. is which consists of n terms, in which the common difference is d. Consider, in the A. P. (a), the terms p and g which are equally distant from the extremes; the first, p, is preceded by three terms and therefore by formula (i) p=a+3d; the second, g, has three terms following it, and, consequently, has three terms preceding it in the A. P. (b), therefore by formula (i) g=1+ 3(— d) = 1—3d; adding, the sum of these two terms is l P+g=a+3d+1-3d= a + 1. In general, consider in the A. P. (a), two terms which are equally distant from the extremes, such that the first has r terms before it, and the second terms following it. By formula (i), the first of these terms is r a +rd; the second can be regarded as a term in the A. P. (b) which is preceded by r terms; it is therefore equal to 1 + r (d) = l—rd; therefore the sum of the two terms is likewise equal to a+rd + (1 - rd) = a + 1. 525. PROBLEM II.-Given the first term and the last term of an A. P.; find the sum of the first ʼn terms. then it follows, on reversing the order of the terms in the second member, that 2 S = (a+1) + (b + k) + (c + h) + (p+g)+... ...+(g+p) + (h+c) + (k + b) + (l +a). are each equal to a + 1; the number of these sums is evidently the same as the number of terms of the A. P. to be summed, that is n; therefore it follows that n (iii) 2 S = n (a + 1) or S = 12 (a + 1). Formula (iii) expresses the sum of the first n terms of A. P. in terms of the number of terms, the first, and the last term. By formula (i) 1 = a + (n − 1) d. 526. Applications.-1. Find the sum of the first n integers These numbers form an A. P.; the first term is 1; the last or nth term is n; and therefore by formula (iii) the sum of the first n terms is S = 22 (n+ 1). 2. Find the sum of the first n odd integers 1, 3, 5, 7, These n numbers form an A. P.; the first term is 1; the common difference is 2; therefore, the sum of the first n odd integers is n = "2" (2 + (n − 1) 2) = 1⁄2) ⋅ 2 n = n2. S= 527. The formulae n 2 give two relations connecting the five quantities a, l, d, n, S; these two equations enable one to calculate any two of these five quantities if the other three are given. There will, therefore, be ten different problems, according as the quantities comprised in one of the following groups are the unknown quantities: (a,1); (a, d); (a, n); (a, S); (l, d); In case a and n, or l and n are taken as the unknown quantities, the problem leads to an equation of the second degree; in all other cases the problem is solved by working out equations of the first degree. EXAMPLE 1. Find the first term and the number of terms in an A. P. in which the common difference is 2, the sum of the series is 72, and the last term is 21. EXAMPLE 2. Sum 100 terms of the A. P, whose third term is 5, 528. PROBLEM III. -Insert n arithmetic means between two given numbers a and b; that is, form an A. P. composed of n + 2 terms, whose extreme terms are a and b. Let the common difference be d, then by formula(i) |