we may immediately perceive the rank which any term holds in the progression. These numbers written above the terms, may be called indices; and the above example is written as follows: Indices, 1 2 3 4 5 6 7 8 9 10 8, 11, 14, 17, 20, 23, 26, 29, &c. Arith. Prog. 2, 5, where we see that 29 is the tenth term. 364. Let a be the first term, and d the difference, the arithmetical progression will go in the following order: a, ad, a +2d, a + 3d, a + 4d, a +5d, a + 6d, &c. whence it appears, that any term of the progression might be easily found, without the necessity of finding all the preceding ones, by means only of the first term a and the difference d. For example, the tenth term will be =a9d, the hundredth term = a + 99 d, and generally, the term n will be = a + (n − 1) d. 365. When we stop at any point of the progression, it is of importance to attend to the first and the last term, since the index of the last will represent the number of terms. If, therefore, the first terma, the difference =d, and the number of terms = n, we shall have the last term = a + (n-1) d, which is consequently found by multiplying the difference by the number of terms, minus one, and adding the first term to that product. Suppose, for example, in an arithmetical progression of a hundred terms, the first term is 4, and the difference = 3; then the last term will be =99 × 3 + 4 = 301. last z, with the 366. When we know the first term a and the number of terms n, we can find the difference d. For, since the last term z = a + (n − 1) d, if we subtract a from both sides, we obtain za (n-1) d. So that by subtracting the first term from the last, we have the product of the difference multiplied by the number of terms minus 1. We have, therefore, only to divide z-a by n 1 to obtain the required value of the difference d, = which will be = za This result furnishes the following rule: Subtract the first term from the last, divide the remainder by the number of terms minus 1, and the quotient will be the difference: by means of which we may write the whole progression. 367. Suppose, for example, that we have an arithmetical progression of nine terms, whose first is = 2, and last 26, and that it is required to find the difference. We must subtract the first term, 2, from the last, 26, and divide the remainder, which is 24, by 91, that is, by 8; the quotient 3 will be equal to the difference required, and whole progression will be 1 2 3 2, 5, 8, 4 5 6 7 8 9 To give another example, let us suppose, that the first term = 1, the last 2, the number of terms = 10, and that the arithmetical progression, answering to these suppositions, is required; we shall immediately have for the difference clude that the progression is 2. 1 1 10 = and thence con9' Another Example. Let the first term = 2, the last = 12}, and the number of terms = 7; the difference will be 2, 4, 5, 75, 9, 10, 121. 368. If now the first term a, the last term z, and the difference d, are given, we may from them find the number of terms n. For a = (n − 1) d, by dividing the two sides by d, we have since z - = n 1. Now, n being greater by 1 than n n =2¬a+ 1; consequently, the number of terms is found by d dividing the difference between the first and the last term, or z — a, by the difference of the progression, and adding unity to the quotient, For example, let the first term 4, the last 100, and the difference 12, the number of terms will be. + 1 = 9; 12 If the first term = 2, the last = 6, and difference = 1, the Again, let the first term = 31, the last = 73, and the difference = 1, the number of terms will be = 72-31 +14; which are, 3, 4, 6, 73. 369. It must be observed, however, that as the number of terms is necessarily an integer, if we had not obtained such a number for n, in the examples of the preceding article, the questions would have been absurd. Whenever we do not obtain an integral number for the value of Za 7, it will be impossible to resolve the question; and conse d quently, in order that questions of this kind may be possible, z — a, must be divisible by d. 370. From what has been said, it may be concluded, that we have always four quantities, or things, to consider in arithmetical progression ; I. The first term a. II. The last term z. III. The difference d. IV. The number of terms n. And the relation of these quantities to each other are such, that if we know three of them, we are able to determine the fourth; for, I. If a, d, and n are known, we have z = a + (n = CHAPTER IV. Of the Summation of Arithmetical Progressions. 371. It is often necessary also to find the sum of an arithmetical progression. This might be done by adding all the terms together; but as the addition would be very tedious, when the progression consisted of a great number of terms, a rule has been devised, by which the sum may be more readily obtained. 372. We shall first consider a particular given progression, such that the first term = 2, the difference 3, the last term = 29, and the number of terms = 10; 1 2 3 4 5 6 7 8 9 10 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. We see, in this progression, that the sum of the first and the last term=31; the sum of the second and the last but one = 31; the sum of the third and the last but two 31, and so on; and thence we conclude that the sum of any two terms equally distant, the one from the first, and the other from the last term, is always equal to the sum of the first and the last term. 373. The reasons of this may be easily traced. For, if we suppose the firsta, the last z, and the difference = d, the sum of the first and the last term is a +z; and the second term being ad, and the last but one = — - d, the sum of these two terms is also =a+z. Further, the third term being a + 2d, and the last but two = Z 2 d, it is evident that these two terms also, when added together make a + z. The demonstration may be easily extended to all the rest. 374. To determine, therefore, the sum of the progression proposed, let us write the same progression term by term, inverted, and add the corresponding terms together, as follows: 2 + 5+ 8 + 11 + 14 + 17 +20 + 23 + 26 +29 292623 +20 +17 + 14 + 11 + 8 + 5 + 2 31 +31 + 31 + 31 + 31 + 31 + 31 + 31 + 31 + 31. This series of equal terms is evidently equal to twice the sum of the given progression; now the number of these equal terms is 10, as in the progression, and their sum, consequently, = 10 X 31 = 310. So that, since this sum is twice the sum of the arithmetical sion, the sum required must be 155. progres 375. If we proceed in the same manner, with respect to any arithmetical progression, the first term of which is = a, the last =z, and the number of terms = n; writing under the given progression the same progression inverted, and adding term to term, we shall have a series of n terms, each of which will be = a+z; the sum of this series will consequently be = n (a + ≈), and it will be twice the sum of the proposed arithmetical progression; which therefore will be = a + 2). n (a 2 376. This result furnishes an easy method of finding the sum of any arithmetical progression; and may be reduced to the following rule: Multiply the sum of the first and the last term by the number of terms, and half the product will be the sum of the whole progression. Or, which amounts to the same, multiply the sum of the first and the last term by half the number of terms. Or, multiply half the sum of the first and the last term by the whole number of terms. Each of these enunciations of the rule will give the sum of the progression. 377. It may be proper to illustrate this rule by some examples. First, let it be required to find the sum of the progression of the natural numbers, 1, 2, 3, &c. to 100. This will be, by the first rule, 100 × 101 2 = 50 X 101 = 5050. If it were required to tell how many strokes a clock strikes in twelve hours; we must add together the numbers 1, 2, 3, &c. as far as 12; now this sum is found immediately If we wished to know the sum of the same progression continued to 1000, we should find it to be 500500; and the sum of this progression continued to 10000, would be 50005000. 378. Another Question. A person buys a horse, on condition that for the first nail he shall pay 5 halfpence, for the second 8, for |