Arithmetical Progressions. Definitions and Principles. 335. Any number of quantities that increase or decrease according to a law constitute a Series, or Progression. 336. The quantities which compose a scries, or progression, are called the terms. 337. A progression in which each term after the first is derived from the preceding term by the addition of a constant quantity, is an arithmetical progression. 338. The constant quantity added to any term of an arithmetical progression to produce the next term, is called the common difference. 339. If the common difference is positive, the series, or progression, is an ascending one; if negative, a descending one. Thus, a, 3a, 5 a, 7a, etc., is an ascending series; and, 7a, 5a, 3 a, a, etc., is a descending series. 340. In the general discussion of arithmetical progressions, a represents the first term, d the common difference, the last term, n the number of terms, and S the sum of the terms. 341. If we represent the first term by a and the common difference by d, the 2d term = a + d 3d term = a + 2 d 4th term = a + 3 d 5th term = a + 4 d Here we observe that each term equals the first term plus the common difference multiplied by the number of terms less one; hence, the nth term = a+ (n-1)d; but the nth term is the last term 7; therefore, Prin. 132. l= a + (n − 1) d. [Formula A.] 342. If we represent the sum by S, we have, S= a + (a + d) + (a + 2 d) + .... If we write the series in an inverse order, 2 S=(1+a) + (l + a) + (l + a) +.... (1 + a) + (l + a)+(l+a); or 2 S=(la) taken n times = (1+a) n; therefore, Prin. 133. S=(1+a). [Formula B.] Examples involving Arithmetical Progressions. Illustrations.-1. Find the last term and sum of the series 3, 7, 11, 15, etc., to 10 terms. = Solution: Here a 3, d= 4, and n = 10. Substitute these values in formula A: Substitute the values of 1, a, and n in formula B: 2. The first term of an arithmetical progression is 25, the number of terms is 6, and the sum of the terms is 102. Required the last term. = Solution Substitute the values a 25, n=6, and S= 102 in formula B: 3. Given 731, d= 4, and S 136, to find n. S=136, Solution Substitute these values in formulas (A) and (B): 1. Since la + (n−1) d, 31 = a + (n-1) 4, or a + 4n = 35 (A) Note. Since the number of terms is a whole number, 8 is the true 1. Find the 10th term and the sum of 10 terms of the series: 4, 8, 12, etc. 2. Find the 12th term and the sum of 12 terms of the series 27, 25, 23, etc. 3. Find the 9th term and the sum of 9 terms of the 4. Find the nth term and the sum of the n terms of the series 1, 2, 3, etc. 5. Find the rth term and the sum of r terms of the series: 2, 4, 6, etc. 6. Given a = 3, 7= = 28, and n = :6, find d. 7. Given S 112, n = 7, and a 25, find 7 and d. = 8. Given n= 8, a = 8, and d = 5, find S and 7. 9. Given d=1, S=58, and a=2, find 1 and n. = 2' 2 (S―an) n (n − 1) 1=61, and S= 45, to find a and n. 2 21. Given d 4, S=190, and a 1, to find 7 and n. 22. Given d= 3, 135, and S=220, to find n and a. 23. Show that n = d-2a√(2a-d)2+8d S Concrete Examples involving Arithmetical Illustrations.-1. Insert three arithmetical means between 3 and 11. Solution: Since there are to be three arithmetical means, the number of terms is 5, the first term is 3, and the last term is 11. |