1 663 + a6 47, &c. If, as before, this series be multiplied by (a + b)2 or aa + 2ab+bb, the product will be found to be =1. 726. If the series found for the value of (a+b) be multiplied by a+b only, the product ought to be the fraction or be equal to the series already found, namely, 63 64 at + a5, 1 &c.; and that, on multiplication, will be found to be the case. ARITHMETICAL RATIO. 1 b bb a a2 + 727. The relation which one quantity bears to another, with respect to magnitude, is called a ratio. It is evident that no relation can exist between quantities that are not of a similar kind; as, for example, a number must be compared with a number, a line with a line, &c. The magnitudes of quantities may be compared in two ways. In the first, they may be compared with regard to their difference; and then the question asked is, how much one quantity is greater or less than another? The relation in this respect, which quantities bear to each other, is called their arithmetical ratio. The other way in which they may be compared is, by inquiring how often one quantity is greater than another? and this relation between quantities is called their geometrical ratio. The term ratio, when simply applied, is generally understood in the latter sense; and we shall reserve the word ratio and relation to express geometrical ratios. 728. By subtraction, the difference is found between two numbers; hence the question, how much the one is greater than the other, is easily resolved. Thus, between two equal numbers, the difference being nothing, if we are asked how much one of the numbers is greater than the other, the answer is, by nothing. Thus, 8 being =2 x 4, the difference between 8 and 2 x 4 is 0. 729. When two numbers, as 5 and 3, are not equal, and we require to know how much 5 is greater than 3, the answer is 2, and it is obtained by subtracting 3 from 5. So 17 is greater than 7 by 10, and 25 exceeds 8 by 17. There are therefore three things relative to the subject for our consideration : 1st, the greater of the two numbers; 2d, the less; and 3d, the difference; which three quantities are so connected, that two of the three being given, the third may also be determined. Suppose the greater number =a, the less =b, and the difference =d, the difference will be found by subtracting b from a, so that d=a—b, whence we find d, if a and b are given. 730. But if the difference and the less b of the two numbers are given, the greater number is determined by adding the difference to the less number, which gives a=b+d; for if we take from b+d the less number b, there remains d, which is the known difference. Let the number 12 and the difference 8, then the greater number = 20. = Lastly, if a the greater be given, and d the difference, b will be found by subtracting the difference from the greater number, that is, b-a-d. 731. The connection, then, among the numbers is of such a nature as to give these results: :- 1st. d-a-b; 2d. a=b+d; 3d. b-a-d; and, generally, if z=x+y, then y=2-x and x=z-y. It must here be remarked, with respect to arithmetical ratios, that if any number, as c, be added to the numbers a and b, the difference is still the same. Thus, d being the difference between a and b, that number d will also be the difference between a + c and b+c, and between a-c and 5-c. Thus, the difference between 20 and 8 being 12, such difference will remain the same whatever numbers we add to 20 and 12, and whatever numbers we subtract from them; for if a-b=d, we must have (a + c) − (b + c)=d, as also (a–c)-(b-c)=d. So, if the numbers be doubled, the difference will be double, and, generally, na-nb=nd, whatever be the value of a. ARITHMETICAL PROPORTION. 732. When two arithmetical ratios or relations are equal, the equality is called an arithmetical proportion. Thus, if a-b-p-q, the difference between p and q being the same as that between a and b, these four numbers are said to form an arithmetical proportion, which is thus written, a-b=p-q. An arithmetical proportion, then, consists of four terms, which are such that, subtracting the second term from the first, the remainder is the same as when we subtract the fourth from the third; thus the numbers 24, 9, 23, 8, form an arithmetical proportion, because 24-9=23-8, which by some is written 24. 9:: 23.8. In any arithmetical proportion, as a−b=p-q, the second and third quantities may change places without changing the equality; for as a-b=p-q, add b to both sides, and we have a=b+p-q; and now subtracting p from both sides, we have a-b-p-q. In numbers, as 24-9=23-8, so 24-23=9-8. In an arithmetical proportion, the second term may take the place of the first, if the fourth be made to take the place of the third; thus, if a-b=p-q, we have b-a=q—p. For b-a is the negative of a-b, and q-p is the negative of p-q. But the great 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; a property which deserves particular consideration, and is expressed by saying that the sum of the means (middle terms) is equal to the sum of the extremes (extreme terms). Thus, since 24-9=23-8, we have 9+ 23 = 24 + 8, both being 32. The demonstration of this is as follows: Let a-b=p-q, add to both b + q, and we have a +q=b+p, that is, the sum of the first and fourth is equal to the sum of the second and third; and inversely, if four numbers, a, b, p, q, be such that the sum of the second and third is equal to that of the first and fourth, that is, if b+p=a+q, we may be sure that those numbers are in arithmetical proportion, and that a-b=p-q; for if a +q=b+p, subtracting from both sides b + q, we obtain a-b-p-q. Thus the numbers 24, 12, 27, 15, being such that the sum of the means (12+27=39) is equal to the sum of the extremes (24 + 15 =39), we are certain that they form an arithmetical proportion, and consequently that 24-12=27-15. 733. By this property, the following question is resolved: — The three first terms of an arithmetical proportion being given, to find the fourth, let a, b, p be the three first terms, and let the fourth, which is that sought, be represented by q. Then a+q=b+p; by subtracting a from both sides we have q=b+p-a. Hence it appears that the fourth term is found by adding together the second and third, and from their sum subtracting the first. Thus, suppose the three first terms are 24, 12, 27, the sum of the second and third is =39, from which subtract 24, the first, and we have 15 for the fourth term sought. When in an arithmetical proportion the second term is equal to the third, we have only three numbers, whose property is this, that the first minus the second is equal to the second minus the third, or that the differences between the first and second and between the second and third are equal. Of this kind are the numbers 23, 18, 13, since 23-18-18-3. Three such numbers, as 23, 18, 13, are said to form a continued arithmetical proportion, called an arithmetical progression, particularly when a great number of such terms follow each other according to the same law. An arithmetical progression may be either increasing or decreasing; that is to say, the former when the terms go on increasing, as 5, 9, 13, and the latter when they go on diminishing, as 12, 9, 6. 734. Let us suppose the numbers a, b, c to be in arithmetical progression; then a-b=b-c; hence from the equality between the sum of the extremes and that of the means, 2b=a+c, if we subtract a from both, we have c=2b-a: hence, when the two first terms a b of an arithmetical progression are given, the third is found by taking the first from twice the second. Thus let 2 and 5 be the two first terms of an arithmetical progression, the third will be 2 × 5-2=8; and the three numbers 2, 5, 8 give the proportion 2-5=5-8. This method enables us to obtain, to any extent, an arithmetical progression; for we have only to find the fourth by means of the second and third in the same way as the third was determined by means of the first and second, and so on. Let a be the first term and b the second, the third will be =2b-a, the fourth 4b-2a-b-3b-2a, the fifth will be 6b-4a-2b+ a=4b-3a, the sixth ·8b-6a −3b+2a=5b-4a, &c. x C ARITHMETICAL PROGRESSION. 735. Having in the preceding subsection seen the nature of arithmetical progression, we may perceive that the natural numbers written in their order (as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, &c.) form an increasing arithmetical progression, because they increase constantly by unity; and the series 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1 is also such a progression wherein the numbers constantly decrease by 2. The number or quantity by which an arithmetical progression increases or decreases is called the difference. Hence, if the first term and difference be given, we may continue it to any extent. For instance, let the first term =3, and the difference =4, we shall have the following increasing progression, 3, 7, 11, 15, 19, 23, 27, 31, 35, &c., wherein each succeeding term is found by adding the difference to the preceding one. It is usual to write the natural numbers 1, 2, 3, 4, &c. above the term of such an arithmetical progression, in order that we may perceive the rank held by any term in the progression. The numbers so written above the terms are called the indices, as in the following example: — Indices 1 2 3 4 5 6 7 8 9 10 11 Arithmetical progression 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, &c. ; 736. Let a be the first term, and d the difference, the following will be the order of it: a, a +d, a + 2d, a + 3d, a + 4d, a +5d, a + 6d, &c. Whence it is evident, that any term of the progression may be found without the necessity of finding the intermediate ones merely by the first term, and the difference d; for example, the twelfth term, =a+11d, the thousandth term, =a+ 999d, and generally the last term =a+ (n-1)d; hence it is only necessary to multiply the difference by the number of terms minus one, and to the product to add the first term. Thus, suppose an arithmetical progression of a hundred terms whose first term is 6, and the difference =5, then the last term will be 99 × 5+6=501. = 737. Knowing the first term a, and the last z, the number of terms », 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 z-a=(n-1)d. Then, by subtracting the first term from the last, we have the product of the difference multiplied by the number of terms minus 1: And dividing z-a by n-1, the required value of the difference will be =; from which results 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 which the whole progression may be written Example. Let the first term = 2, the last 26, to find the difference, - the quotient = 3 will be equal to the difference, and the progression will 738. Another example. then the difference will be 1 2 3 4 5 6 7 8 9 2, 5, 8, 11, 14, 17, 20, 23, 26. Let the first term = 24, the last =12, the number of terms =7; 21, 45, 7, 9, 10, 121. 739. If the first term a, the last term z, and the difference d be given, we may from these find the number of terms n. For, inasmuch as z-a=(n−1)d, if we divide the two sides by d, we have "=n-1, and n being greater by 1 than n-1, we have n= d n=d+]; the number of terms is therefore found by dividing the difference between the first and last terms, or z-a by the difference of the progression, and adding unity to the quotient *~. 740. Thus, for example, let the first term =4, the last =100, and the difference=12; the number of terms will be +1=9, and these nine terms will be 100-4 12 1 2 3 4 5 6 7 8 9 4, 16, 28, 40, 52, 64, 76, 88, 100. Another example.-Let the first term =3}, the last=73, and the difference=1; the number 73-31 of terms will be +1=4, which are 31, 47, 64, 73. 13 741. It must, however, be remarked, that the number of terms being necessarily an integer, if such a number had not been obtained for n in the foregoing examples, the questions would have been absurd; and if an integer number be not obtained for the value , the question cannot be resolved; hence, in order that such questions may be possible, z-a must be divisible by d. 742. It may now be concluded, then, that there are always four quantities for consideration in an arithmetical progression. 1. The first term, a. 2. The last term, z. 3. The difference, d. 4. The number of terms, n; and the relations of these to each other are such, that, if we know three of them, the fourth may be found. For, 1. If a, d, and n are known, we have z=a+ (n−1)d. 2. If z, d, and n are known, we have a=2— — (n−1) d. 3. If a, z, and n are known, we have d=" 4. If a, z, and d are known, we have 2-a SUMMATION OF ARITHMETICAL PROGRESSIONS. 743. To find the form of an arithmetical progression by adding all the terms together would be troublesome when the number of terms is very great; a rule has therefore been found by which the operation is much shortened. Let us first consider a particular given progression, whose first term =2, difference =3, the last term =29, and the number of terms 10. 1 2 3 4 5 6 7 8 9 10 744. In this progression, the sum of the first and last term =31, the sum of the second and last but one =31, and so on; from which we conclude that the sum of any two terms equally distant, the one from the first, the other from the last term, is always equal to that of the first and last. It will not be difficult to discover the cause of this; for, suppose the first =a, the last =z, and the difference =d, the sum of the first and last is a+z, and the second term being =a+d, and the last but one =z-d; the sum of these two terms is also =α +2. Again, the third term being a + 2d, and the last term but two =2=2d, it is evident that the sum of these two terms also makes a + z. From this, the demonstration for the rest is obvious. Now, if we write the progression term by term twice over, but in one line, invert the order of the terms, and add the corresponding terms together, we shall have as follows: 2+ 5+ 8+11 +14 +17 +20 +23+ 26+ 29 29 +26+ 23 +20 +17 +14 +11+ 8 + 5+ 2 31 +31 +31 +31 +31 +31 +31 +31 +31 + 31 A series of equal terms, evidently equal to twice the sum of the given progression, whose number of terms being 10, the sum here exhibited will be = 10 x 31=310. Hence, as this is twice the sum of the arithmetical progression, the sum required must be 155. 2 745. Treating in the same manner any arithmetical progression whose first term = a, last term =z, and number of terms =n, and writing as above shown, the progression direct and inverted, the one under the other, and adding term to term, we have a series or n terms each =a+z, whose sum will consequently be =n(a + z), which will be twice the sum of the proposed arithmetical progression, and therefore =(a+), From which flows a simple rule for finding the sum of an arithmetical progression. Multiply the sum of the first and last terms, and half the product will be the sum of the whole progression. We will illustrate this rule by an example. Let it be required to find the sum of the progression of the natural numbers, 1, 2, 3, &c. to 100. This, by the rule, will be 100 x 101-50 x 101 =5050. 2 103 × 32 746. Let it be required to find the sum of an arithmetical progression whose first term is 5, the difference =3, and the number of terms 32: we must begin by using the rules in subsections 735. et seq., by which we determine the last term to be =5+31 x 3=98, after which the sum is immediately seen to be =103 × 16 =1648. Generally, to find the sum of the whole progression, let the first term be =a, the differenced, and the number of terms =n. Now, as by the preceding subsection the last term must be = a + (n−1)d, the sum of the first and last will be 2a + (n-1) d; multiplying this sum by the number of terms n, we have 2na+n(n−1)d; the sum required, therefore, will be na + (n-1), and this formula, applied to the preceding example, gives 1648, as before. 2 747. Suppose it required to add together all the natural numbers from 1 to n, we have for resolving the question the first term =1, the last term =n, and the number of terms =n, the sum required is Let n be 1766, then the sum of all the num = nn+n_n(n+1) 2 2 bers from 1 to 1766-883 × 1767 1560261. 748. If a progression of uneven numbers be proposed, 1, 3, 5, 7, &c. continued to n terms, and the sum be required. We have the first term =1, the difference 2, the number of terms =n; the last term will therefore be =1+(n−1)2=2n−1, and, consequently, the sum = nn. Hence, whatever number of terms of this progression are added together, the sum will always be a square, namely, the square of the number of terms, as a view of the following table will render manifest: Indices 1 2 3 4 5 6 7 8 9 10, &c. - 1 3 5 7 9 11 13 15 17 19, &c. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, &c. The subjoined table exhibits formula for differences up to 10: 749. We have before observed that the geometrical ratio of two numbers is found by answering the question how many times is one of those numbers greater than the other, the quotient being the ratio required. Three things here present themselves for consideration. Firstly, the first of the two given numbers, which is called the antecedent; secondly, the other number, which is called the consequent; thirdly, the ratio of the two numbers, or quotient arising from the division of the antecedent by the consequent. Thus, if the relation of the numbers 18 and 12 be sought, 18 being the antecedent, and 12 the consequent, the ratio will be 1=1, whence we see the antecedent contains the consequent once and a half. Geometrical relations are generally represented by a point placed above another between the antecedent and the consequent. Thus ab signifies the geometrical relation of these two numbers, or the ratio of b to a. The sign just mentioned has, in a previous page, been given as indicating division, and it is on that account here used, because, in order to know the ratio, we divide a by b. The relation is merely read a is to b. On this account relation is expressed by a fraction whose numerator is the antecedent and denominator the consequent. It is hardly necessary to say that this fraction should, for perspicuity sake, appear in its lowest terms. Thus, if both terms be divided by 6, becomes }. 750. Hence relations differ as their ratios are different; and there are, of course, as many kinds of geometrical relations as we can imagine different ratios. The first kind is that wherein the ratio becomes unity, which, of course, happens when the two numbers are equal, as in 3: 3, 4: 4; a; a, and because the ratio here is 1, it is called the relation of equality. The relation then follows in which the ratio is another whole number; in 4 : 2 the ratio is 2, and is called a double ratio; in 12: 4, the ratio being 3, it is called a triple ratio; in 24 6 the ratio is 4, and is called a quadruple ratio. It is necessary, also, to notice those relations whose ratios are expressed by fractions, as 12: 9, where the ratio is , or 1; 18: 27, where the ratio is 3, &c. Those relations, as 6: 12, 5; 15, &c., wherein the consequent contains exactly twice, thrice, &c. the antecedent, are sometimes called subduple, subtriple, &c. ratios. The term rational is applied to ratios that are expressible numbers, the antecedent and consequent being integers, as 11: 7, 8: 15, &c.; and that of irrational, or surd, is applied to ratios neither expressible by integers nor by fractions, 58, 4: √3. as a 751. If a be the antecedent, b the consequent, and d the ratio, d=%. Were the consequent b given with the ratio, we should find a=bd, for bd divided by b gives d. Finally, when the antecedent a is given, and the ratio d the consequent, b=; for, dividing the antecedent a by the consequent (or its equivalent), we obtain the quotient d, that is, the ratio. In whatever way we multiply or divide the consequent and antecedent by the same number, every relation ab remains the same, because the ratio is the same. Let d be the ratio of ab, we have then d=: : now the relation na ; nb is still =d, and likewise a a is still =d. So, also, when a ratio has been reduced to its lowest terms, the relation is easily perceived and enunciated; for, let the ratio be reduced to its lowest terms?, we say, a b=p: q, ab::p: q, which is read a is to b as p is to q. Thus the ratio of the relation 6 3 being, or 2, we say, 6:3=2; 1. So, 18 12 3: 2, and 2418: =43, &c. But if the ratio be not susceptible of abridgment, the relation does not become more evident, and we do not simplify a relation by saying 9:79: 7... We may, however, change the relation of two large numbers into one more simple and evident by reducing both to their lowest terms, for we may say, 14484: 7242=2:1, or 15819 10546=3 2, or 57600; 25200=16; 7. All relations, therefore, should be reduced to the lowest possible numbers, which is readily done by dividing the two terms of the relation by their greatest common divisor. Thus, to reduce the relation 57600 25200 to that of 16 7, we have only to perform the single operation of dividing the numbers 576 and 252 by their greatest common divisor, 36. The method of finding a common divisor of two given numbers will be given in the following subsection. |