The material of this work is aimed at mathematics educators, as well as math specialists with a keen interest in progressions. In this paper, we study the subject of arithmetic, geometric, mixed, and harmonic progressions or sequences. Some of the material found in Sections 2,3,4,and 5, can also be found in standard precalculus texts. For example, refer to the books in references 1 and 2. A substantial portion of the material in those sections, cannot be found in such books. In Section 6, we present 21 problems with detailed solutions. These are interesting, unusual problems not commonly found in mathematics texts, and most of them are quite challenging. In Section 7, we offer a collection of unsolved problems.
Deep Dive into Exploring Progressions: A Collection of Problems.
The material of this work is aimed at mathematics educators, as well as math specialists with a keen interest in progressions. In this paper, we study the subject of arithmetic, geometric, mixed, and harmonic progressions or sequences. Some of the material found in Sections 2,3,4,and 5, can also be found in standard precalculus texts. For example, refer to the books in references 1 and 2. A substantial portion of the material in those sections, cannot be found in such books. In Section 6, we present 21 problems with detailed solutions. These are interesting, unusual problems not commonly found in mathematics texts, and most of them are quite challenging. In Section 7, we offer a collection of unsolved problems.
In this work, we study the subject of arithmetic, geometric, mixed, and harmonic progressions. Some of the material found in Sections 2,3,4, and 5, can be found in standard precalculus texts. For example, refer to the books in [1] and [2]. A substantial portion of the material in those sections cannot be found in such books. In Section 6, we present 21 problems, with detailed solutions. These are interesting, unusual problems not commonly found in mathematics texts, and most of them are quite challenging. The material of this paper is aimed at mathematics educators as well as math specialists with a keen interest in progressions.
In this paper we will study arithmetic and geometric progressions, as well as mixed progressions. All three kinds of progressions are examples of sequences. Almost every student who has studied mathematics, at least through a first calculus course, has come across the concept of sequences. Such a student has usually seen some examples of sequences so the reader of this book has quite likely at least some informal understanding of what the term sequence means. We start with a formal definition of the term sequence.
(a) A finite sequence of k elements, (k a fixed positive integer) and whose terms are real numbers, is a mapping f from the set {1, 2, . . . , k} (the set containing the first k positive integers) to the set of real numbers R. Such a sequence is usually denoted by a 1 , . . . , a n , . . . , a k . If n is a positive integer between 1 and k, the nth term a n , is simply the value of the function f at n; a n = f (n).
(b) An infinite sequence whose terms are real numbers, is a mapping f from the set of positive integers or natural numbers to the set of real numbers R, we write F :
Such a sequence is usually denoted by a 1 , a 2 , . . . a n , . . . . The term a n is called the nth term of the sequence and it is simply the value of the function at n. Remark 1: Unlike sets, for which the order in which their elements do not matter, in a sequence the order in which the elements are listed does matter and makes a particular sequence unique. For example, the sequences 1, 8, 10, and 8, 10, 1 are regarded as different sequences. In the first case we have a function f from {1, 2, 3} to R defined as follows: f := {1, 2, 3} → R; f (1) = 1 = a 1 , f (2) = 8 = a 2 , and f (3) = 10 = a 3 . In the second case, we have a function g : {1, 2, 3} → R; g(1) = b 1 = 8, g(2) = b 2 = 10, and
Only if two sequences are equal as functions, are they regarded one and the same sequence.
Definition 2: A sequence a 1 , a 2 , . . . , a n , . . . with at least two terms, is called an arithmetic progression, if, and only if there exists a (fixed) real number d such that a n+1 = a n +d, for every natural number n, if the sequence is infinite.
If the sequence if finite with k terms, then a n+1 = a n + d for n = 1, . . . , k -1. The real number d is called the difference of the arithmetic progression.
Remark 2: What the above definition really says, is that starting with the second term a 2 , each term of the sequence is equal to the sum of the previous term plus the fixed number d.
Definition 3: An arithmetic progression is said to be increasing if the real number d (in Definition 2) is positive, and decreasing if the real number d is negative, and constant if d = 0.
Remark 3: Obviously, if d > 0, each term will be greater than the previous term, while if d < 0, each term will be smaller than the previous one.
Theorem 1: Let a 1 , a 2 , . . . , a n , . . . be an arithmetic progression with difference d, m and n any natural numbers with m < n. The following hold true:
(i) a n = a 1 + (n -1)d (ii) a n = a n-m + md (iii) a m+1 + a n-m = a 1 + a n Proof:
(i) We may proceed by mathematical induction. The statement obviously holds for n = 1 since a 1 = a 1 +(1-1)d; a 1 = a 1 +0, which is true. Next we show that if the statement holds for some natural number t, then this assumption implies that the statement must also hold for (t + 1). Indeed, if the statement holds for n = t, then we have a t = a 1 +(t-1)d, but we also know that a t+1 = a t + d, since a t and a t+1 are successive terms of the given arithmetic progression. Thus,
, which proves that the statement also holds for n = t + 1. The induction process is complete.
(ii) By part (i) we have established that a n = a 1 + (n -1)d, for every natural number n. So that a n = a 1 + (n -1)d -md + md;
Again, by part (i) we know that a n-m = a 1 + [(n-m) -1]d. Combining this with the last equation we obtain, a n = a n-m + md, and the proof is complete.
(iii) By part (i) we know that a m+1 = a 1 +[(m+1)-1]d ⇒ a m+1 = a 1 +md; and by part (ii), we have already established that a n = a n-m + md. Hence, a m+1 + a n-m = a 1 + md + a n-m = a 1 + a n , and the proof is complete.
Remark 4: Note that what Theorem 1(iii) really says is that in an arithmetic progression a 1 , . . . , a n with a 1 being the first term and a n being the nth or last term; if we pick
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