304. In any arithmetical proportion we may change the places of the second and third terms, that is to say, if a-bp-q, then will For if a-bp-q, adding b to each side of the equation, a=b+p-q. Again, subtracting p, a-p=b-q. Thus, if 12-79-4, then will 12-9-7-4. We may also place the second term in place of the first, if we make the same transposition of the fourth and third; that is, if a-bp-q, then will 305. But the principal property of every arithmetical proportion is this; that the sum of the second and third terms is always equal to the sum of the first and fourth, or, in other words, the sum of the means is equal to the sum of the extremes. Thus, if 12-7-9-4, then will 12+4=7+9; for, if a-bp-q, adding b+q to each side, a+q=b+p. Reciprocally also, if a+q=b+p, then will a-b-p-q. That is, if the four quantities a, b, p, q, are such that the sum of the first and last be equal to the sum of the second and third, then do those quantities form an arithmetical proportion. For if a+q=b+p, subtracting b+q from each side, a-bp-q. Thus the numbers 18, 13, 15, 10, being such that 13+15= 18+10, we conclude with certainty that these numbers form an arithmetical proportion, and that 18—13—15–10. 306. By 306. By means of this property we are therefore at no difficulty in determining how, from three given numbers in arithmetical proportion, to find the fourth. Let a, b, p be the three first terms, and let the fourth be represented by q, we are then certain that From which we infer that the fourth term is found by subtracting the first from the sum of the second and third. Suppose, for example, that 19, 28, and 13 were given as the three first terms, to find the fourth. Here q=28+ 13 − 19 = 22. The proportion will therefore be, 19-2813 – 22, or 28-19-22 — 13, or 28-2219-13. 307. In any arithmetical proportion, if the second term be equal to the third, we have in fact only three numbers, but with this property, that the first minus the second is equal to the second minus the third. For example, in the three numbers 19, 15, 11, 19 15 15-11. or generally, abb―q. Three numbers of this sort are said to form a continual arithmetical proportion, which is usually written thus, 19: 15 :11. Such proportions are also called arithmetical progressions, particularly if a greater number of terms follow each other according to the same law. An arithmetical progression may be either increasing or decreasing. The former distinction is applied when the terms go on increasing, that is to say, when the second exceeds the first, the third the second, the fourth the third, &c. by the the same quantity; as in the numbers 4, 7, 10; and the decreasing progression is that in which the terms go on continually diminishing by the same quantity, as 9, 7, 5, 3, 1. 308. Let us suppose the numbers a, b, c, to be in arithmetical progression: then a-b-b-c; whence it follows, from the equality between the sum of the means and that of the extremes, that 2ba+c, or 2b-a-c. So that when the first two terms of an arithmetical progression are given, the third may be found by subtracting the first from twice the second. For instance, let 1 and 3 be the first two terms of an arithmetical progression, the third will be 6-1=5, and these three numbers give the proportion, 1-3 3-5. 309. By following the same course we pursue the progression as far as we please, for we have only to find the fourth term by means of the second and third, in the same manner as we determined the third by means of the second and first, and so on. Let a and b be the two first terms, then the Thus, if it were required to find the 20th term of the progression, 5, 8, 11, &c., we shall have n=20, b =8, a=5, consequently, (n−1).b— (n—2), a = 19 × 8 − 18 x 5152-90=62. CHAP. XXXIX. OF ARITHMETICAL PROGRESSION. 310. WE have already observed that a series of numbers, consisting of any number of terms which continually increase or diminish by the same quantity, is termed, Arithmetical Progression. Thus the natural numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, &c. form an arithmetical progression, since they continually increase by unity; and the contrary series, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, are equally in arithmetical progression, because they continually diminish by the same quantity. 311. The number or quantity by which the several terms of an arithmetical progression are increased or diminished, is termed the common difference. So that when the first term and the common difference are known, we may continue the progression to any number of terms. For example, let the first term = 2, and the common difference 3, we shall have the increasing progression, 2, 5, 8, 11, 14, &c., where each succeeding term is found by adding 3 to the preceding one. 312. Now assuming a as the first term of an arithmetical progression, and b for the common difference, the series will be continued in the following order, a, a+b, a+2b, a+3b, a+4b, a+5b, a+6b, &c. from which it is evident that any one of these terms might at once be found, without the necessity of finding all the preceding ones, and that by means only of the first term a and the common difference b; for, by attending to the above series, it will be seen that the coefficient of b, the common difference, is always less by unity than the number which denotes its place in the series, the numbers 1, 2, 3, 4, &c. written over each term of the series, being denominated the indices of the progression ; thus in the 2d there is a+b, We know therefore that the 10th term will be a +9b, and generally that the nth term will be a + (n-1). b. 313. When we stop at any point of the progression, it is important to attend to the first and last term, since the index of the last term will denote the number of terms. If therefore the first term be a, the difference b, and the number of terms n, we shall have for the last or nth term, a ± (n−1). b, which is consequently found by multiplying the common difference by the number of terms minus 1, and adding or subtracting that product from the first term. Suppose, therefore, in an increasing arithmetical progression of a hundred terms, that the first term were 4 and the difference 3; then will the last term be (100-1). 3+4=301. Or if in a decreasing progression of terms the first term were 13 and the common difference 2, then will the last term or 1 = a − (n − 1). b = 13—6.2 — 13 – 12 = 1, and the series will then be 13, 11, 9, 7, 5, 3, 1. 314. Again, if the first term a, and the last 7, be given, b the common difference may be found; for since the last term, or l,=a±(n−1).b, then will l—a—(n−1). b, and consel-a quently d From hence we obtain the following n-1' general rule, viz. that the common difference is equal to the last term minus the first divided by the number of terms minus 1. Suppose, |