ber of the term a + (n 1)d, (a), in which n denotes the num 1)d. It is easy to see that we may represent any term of the series (a) by a + (n-1)d; for if we put n = 1, a + (n · 1, a + (n − 1)d becomes a, the first term of the series; and if we put n = = 2, a + (n-1)d becomes a + d, and if n = 3, a + (n-1)d becomes c + 2d, and so on to any extent; hence, since any term of (a) results from a + (n − 1)d by putting n equal to the number of the term, a + (n − 1)d is called the general term of the series. (7.) Again, if in (a) we put d for+d, we shall have the decreasing arithmetical progression a, ad, ad, a - 2d, a 2d, a - 3d, a - 4d, a - 5d, . . . . a − (n − 1)d, (a1), whose general term is expressed by a − (n - 1)d. (8.) It is easy to see that we may write the series (a) and (a1) in one series, by writing a, a±d, a ± 2d, a ± 3d, a ± 4d, a ± 5d, . . . . a ± (n − 1)d, (A); observing that for we must use when the series is increasing, and for ± we must use when the series is decreasing. .... (9.) It is easy to see that in the series (A) we have to consider a, which is called the first term, and a (n-1)d, which is called the last term, and n which is supposed to denote the number of terms, together with ±d which is called the arithmetical ratio, or the common difference of any two successive terms; and, in addition to these, it is often necessary to consider the sum of all the terms. (10.) If we represent the last term by 7, we shall have l = a ± (n − 1)d, (1), for the equation that connects the first term, last term, common difference, and number of terms with each other; observing that for we must use + when the series is increasing, and must be used for when the series is decreasing. (11.) If we know a, the first term, the common difference d, and the number of terms n, it is easy to find l, the last or nth term from (1); for it requires us to multiply the common differenced by the number of terms less one, and to add the first term to the product (according to their signs) for the last term. Thus, if the first term a = 2, the common difference = 3, and the number of terms n = 7; (1) gives 7 = 2 + (7 − 1) × 3 = 2 + 6 × 3 = 20 = the last or seventh term, which is clearly correct, for the series is 2, 5, 8, 11, 14, 17, 20, 23, etc., whose seventh term is 20, as it ought to be. Again, if we put a = 40, d=5, n = 8, we have, from (1), 40 (81) x 540-7 x 55 the last = · term, as required; for the series is 40, 35, 30, 25, 20, 15, 10, 5, 0, — 5, — 10, -15, etc., whose eighth term is 5. .... (12.) Since the last term of the series (A) is represented by l, it is easy to see that the last term but one will be expressed by ld, and the last term but two will be expressed by 2d, and so on, to the n" term, which will be expressed by (n-1)d. Hence, if we add the sum of the series. (A) to the sum of the same series when taken in a reverse order, we shall have a + (a + d) + (a ± 2d) + .... +(a± (n − 1)d) + l + (l + d) + (l + 2d) + + (l + (n. 1)d) =na+nl; since the terms that involve d destroy each other on account of the opposition of signs, and that there are n terms in the series (A), so that a and 7 must each be taken n times in the sum. Hence, if s denotes the sum of the series (A), then 28 will equal the sum of the series (A) together with the sum of the same series taken in reverse order; consequently we shall have 2s equal to na + nl = n(a + 1), and n(a + 1) dividing these equals by 2, we have 8 = 2 , (2). (13.) (2) is the equation that connects 8, n, a, and 7; and shows that in order to find the sum of a series of numbers in arithmetical progression, we must take one half of the product of the sum of the extremes (or of the first and last terms added according to their signs) and the number of terms. Thus, if a = 13, 7 = 119, n= 40, we have a + 1 = 132, 40 x 132 =2640, as required. and s = n(a + 1) 2 becomes & 3, ± d = 4, + 20 × 4 = 83; n(a + 1) 21 × (383) 2 = 2 2 n = 21, we get by (1) 7 = a ± and then by (2) we shall have = 903, which is the sum of 21 terms of the series whose first term is 3 and ratio of increase 4. For another example, we shall find the sum of the series 12, 10, 8, 6, 4, 2, 0, 12 and last term is 2, 4, 6, 8, whose first term is 8, and number of terins 11. Here a = 12, l = — 8, n = 11, and by (2) we shall have s = n(a + 1) 2 = 11(12-8) = 22; which is easily seen to be cor rect, by actually adding the terms of the given series according to their signs. (14.) If a and 7 together with n are known, then we may findd, and thence determine the terms of the series that are situated between the first term a and last term 1, which is called the method of finding means between given extremes, or intercalating deficient terms. For, since (1) gives l = a ± (n − 1)d, we easily find ± d a = n1, (3), which is evidently true, from the law of formation 1' of the terms of an arithmetical progression. Since n denotes the number of terms in the series, n - 1 is one less than the number of terms, or one more than the number of terms that are situated between the extremes; hence it follows from (3), that if we divide the difference between the extremes by the number of terms to be intercalated increased by 1, the quotient will be the common difference of any two successive terms. Thus, if a 7, 122, n = 6, we have id= = 5 the number of means to be d= 3, we have a ±d=7+ 310, a2d 13, a 3d 16, a 4d 19, for the ± terms that are to be intercalated between 7 and 22; and we shall have 7, 10, 13, 16, 19, 22 for the series, when it is completed by intercalating the terms that have been found. For another example, we shall suppose that seven terms are to be intercalated between 12 and 12. Here a = 12, 7 = — 12, and n−1=7+1= 8; and we shall have ±d= 12+12 8 a = n = 3. Hence, as in the last example, we get 12, 9, 6, 3, 0, -3, -6, 9, 12 for the series as completed by intercalating the seven terms, as required. (15.) Because a, d, l, n, and s are so connected with the arithmetical progression that its terms may be made to de pend upon them, they are often called its conditions or terms. And since a, l, n, and ±d are connected by (1), and s, n, a, and I are connected by (2), it is evident that any two of the above five terms may be found from the solution of equations (1) and (2), in terms of the other three; so that only three of the above terms can be assumed at will; in other words, only three of the above five terms can be regarded as independent of each other. (16.) It is easy to see from (A), that any three successive terms of an arithmetical progression are in arithmetical proportion; so that we may regard an arithmetical progression as being a continued arithmetical proportion. (17.) It follows, also, from (A), if we take any three terms of an arithmetical progression, such that the intermediate term is equidistant from the other two, that the three terms will be in arithmetical proportion, so that the sum of the extremes is equal to twice the mean. Thus, if we add the second and eighth terms of (A), the sum will equal twice the fifth term. It is also clear that if we take four terms of an arithmetical progression, such that the intermediate terms are equidistant from the corresponding extreme terms, that the sum of the extreme terms equals the sum of the two mean terms. Thus, if we add the second and ninth terms of (A) for the extremes, and the fifth and sixth terms which are equidistant from them for the means, we shall find that the sum of the extremes equals the sum of the means. (18.) Again, if we have the arithmetical proportion a-b = bc, then, by the definition of an arithmetical progression, a, b, c will be three successive terms of an arithmetical progression. Thus, since 2016 1612, it follows that 20, 16, 12 are three successive terms in an arithmetical progression, whose ratio of decrease is 4; and it is easy to continue the progression to any extent. = (19.) But if we have the arithmetical proportion a − b = с ed, then a, b, c, d can not be in arithmetical progression if ab is not equal to be, and the proportion is said to be discontinued; so that a, b, c, d are said to be in discontinued arithmetical proportion. Thus, 12, 10, 6, 4 are in discontinued arithmetical proportion. (20.) If, however, we have a bed, and at the same time a - bbc, we shall of course have ab= b — c = cd; and a, b, c, d will be in continued proportion, or they will be four successive terms of an arithmetical progression. Thus, 12, 10, 8, 6 are in continued arithmetical proportion, and they are four successive terms of an arithmetical progression, whose ratio of decrease is 2. (21.) Finally, it may be proper to notice that in an arithmetical proportion, the first and third terms are sometimes called the antecedents, the first term being called the first antecedent, and the third term being called the second antecedent; also the second and fourth terms are called the consequents, the second term being called the first consequent, and the fourth term the second consequent. Thus, in the proportion 12-10 = 24— 22, 12 is the first antecedent, and 10, that is subtracted from it, is the first consequent; 24 is the second antecedent, and 22, which is taken from it, is the second consequent. QUESTIONS FOR EXERCISE. 1. Find a third arithmetical proportional to 13 and 16. Ans. 16 x 2 — 13 = 19. 2. Find a fourth arithmetical proportional to 27, 33, 78. Ans. 33+ 78 — 27 84. 3. Find an arithmetical mean between 62 and 92. = 5. Find an arithmetical mean between a and c. 6. The first term of an arithmetical progression is 5, the ratio of increase 3; to find the eleventh term. Ans. 5+ 10 × 3 = 35. 7. The first term of an arithmetical progression is 25, and the last term is 125, and the number of terms is 50; to find 25 + 125 the sum of all the terms. Ans. 2 × 50 3750. |