h. When x cy, then also x oc my, or x:x::my : my' ; . . x : my :: x': my'. Consequently if m is so assumed that x=my, then in all cases x'=my'. Whence, if x vary as y, x is equal to y multiplied by some constant quantity. In general, if X, Y are two corresponding values of x, y, X=mY. X Whence, since from X=mY is obtained m= it follows that when two corresponding values of x, y are known the constant m can be determined. k. Let x ccy, and let m denote any exponent, integer or fractional. Since xy, x: x':: y: y'; x": x'" :: y": y'"; that is, " o ym. Whence, if one quantity vary as another, any power or root of the first quantity varies as the same power or root of the second quantity. 1. Let x oy, and let t be another quantity, either variable or constant, and of which t, t' are either different or equal values; Then since x ∞ y, x: x' :: y: y', and tttt (whether t is variable or constant), xt: x't'::yt y't', or xt ∞ yt, y y' x : t t ť" or t t Whence, if one quantity vary as another, and if each of them be multiplied or divided by any quantity, variable or constant, the products or quotients will vary as each other. m. Let xy ∞ X, that is, let xy: x'y' :: X : X'; X y Whence, if two quantities jointly vary as a third quantity, each of the two quantities varies as the third directly and the other of the two inversely. n. If X=X'=constant. ry: 1 :: x'y' : 1. Whence, if the product of two variable quantities is constant, these quantities vary inversely as each other. o. Let a be a constant and x, y, z variables, and let a:x::yz, a : x':; y': z', &c. Consequently if four quantities are always proportional, and one or two of them are constant, the others being variable, it can be found how the latter vary. p. Let x, y, z be three quantities, of which x ∞ y when z is constant and x ∞ z when y is constant ; it is required to determine the variation of x when y, z are both variable. Suppose, first, that r is made to vary as y, and that when y becomes y', x becomes x', DD Next, that x' (varied from x by the variation of y) is made further to vary as z, and that when z becomes z', x' becomes x'. Then since x: x' :: y: y', and 'x'::z: z'; .'. xx' : x'x'::yz: y'z', Therefore if a vary as y when z is constant, and as z when y is constant, when yz are both variable a varies as the product yz. x Similarly, it can be proved that if t varies as v, x, y, z singly, the others being constant, when v, x, y, z are all variable t varies as the product vxyz. OF ARITHMETICAL PROGRESSION. 183. If the continued arithmetical proportion a-b-b-c is prolonged indefinitely, the series of equidifferences composes an arithmetical progression, or progression by difference. An arithmetical progression is, consequently, a series composed of an indefinite number of terms, such that the difference of any two consecutive terms is a constant quantity. The constant difference, when the series is increasing, is additive; and when the series is decreasing, subtractive. This difference is termed the ratio of the progression. The notation a-b-b-c-c-d=d-e is replaced by which has the same meaning. ÷ 3. 7. 11. 15. 19. 23. 27. 31, &c. 45. 41. 37. 33.29. 25. 21. 17, &c. are examples of arithmetical progressions. In the first the ratio is 4, and in the second -4. The first is an increasing, the second a decreasing series. 184. Leta.b.c.d.e.... be an arithmetical progression whose ratio is . Then, since by the definition each term is equal to the preceding term plus the ratio, b=a+&, c=b+c=a+2&, d=c+d=a+3ê, &c. Therefore the corresponding terms of the series ÷ a. b. с a.a+d.a+28. a+38. a+48 are equal. Since the second term=a+1xò. (1=2−1), the fourth term=a+3×8 . (3—4—1); it follows that the nth term (which may be denoted by l, and which has n−1 terms before it,) is expressed by l=a+(n−1); and generally, that any term whatever is equal to the first term plus as many times the ratio as there are terms before it. Since the expression la+(n-1) is changed into the first term by making n=1, into the second by making n=2, into the third by making n=3, &c., it is called the general term. By substituting for & in l, the general term of a decreasing series is obtained; viz. l—a—(n−1)♪. The formula la+(n−1), expressing a relation between a, d, n, l, is sufficient for the resolution of questions in which any three of these quantities are given to find the fourth. For example, if a, l, n are given to find d. Since la+(n−1)d, 1-a - 1. Whence, to find the ratio, it is necessary to subtract the first term from the last, and to divide the remainder by the number of mean terms augmented by 1. As an illustration of the use of this formula, let it be required to insert m arithmetical means between the two given numbers a, l; in other words, to find m quantities comprehended between a, l, and, forming with a, l, an arithmetical progression beginning with a and ending with 7. In this question, the given quantities are, the first term a, the last term 7; the number of terms which, including a and 7, is m+2, and the quantity which is required is the ratio ; for when ô is known, the other terms of the progression can be formed by successive additions. In formula 1, n denotes the number of terms, making n=m+2, and by consequence n−1=m+1. Whence the m+2 terms of the series beginning with 2 and ending with 17 are 2.5.8.11. 14. 17. This particular case is a solution of the problem; to insert four arithmetical means between 2 and 17. It is to be inferred, from the preceding example and conclusion, that, an arithmetical progression being given, if the same number of arithmetical means is inserted between each term and the following, the new series will still be an arithmetical progression. For partial progressions will then be formed, of which all will have the same ratio; and as the last term of each successive partial progression is also the first term of the following progression, the whole of these partial progressions, taken in order, must form an arithmetical progression. 185. Let it be next required to find the sum of the terms of an arithmetical progression + a. b. c. d. i. k. l, the number of terms being n, and the ratio d. Denoting the required sum by S, and arranging the terms consecutively, first, from a to 1, second from 1 to a, Adding, 2 S=(a+1)+(b+k)+(c+i)+(d+h)+ &c. Now, a+c=b, l—i=k .'.(a+ô)+(?—¿)=b+k or a+l=b+k, b+d=c, k—d=i .'.(b+ô)+(k—¿)=c+i or b+k=c+i, c+c=d, i—c=h .'.(c+c)+(i−c)=d+h or c+i=d+h, Wherefore a+i=b+k=c+i=d+h= ... that is, the sum of any two terms, which are equally distant from the extremes, is equal to the sum of the extremes. Consequently, the sum of all the terms, in both series, is equal to the sum of the extremes as often repeated as there are terms in one series. The sum of the extremes being a+l, and the number of terms n, the sum of all the terms in the two equal series is (a+n. The sum of all the terms of the progression a. b. c.d..... i.k.l is, therefore, (a+1)n 2 (a+1)n Therefore S= 2 2. In words, the sum of the terms of an arithmetical progression is equal to half the sum of the extreme terms multiplied by the number of terms. 186. Five quantities, a, d, l, n, S, enter into the formula. Whence, as two equations are sufficient for the discovery of two unknown quantities, it is possible to find any two of the quantities a, &, l, n, S, when the three others are given. Formulæ 1, 2 give, therefore, the solutions of as many distinct problems as there are combinations of five quantities, taken two by two. The number of such combinations is 5X4 =10. In order that these problems may be possible, it is necessary that the value of ʼn be not only real, but also integer and positive, for the number of terms can neither be fractional nor negative. The ten problems, with their solutions, are, l-a 1st. Given a, d, n, to find l, S?.Ans. l=a+(n−1)d, S=1n[2a+(n−1)ĉ]. 2d. Given l, &, n to find a, S?.Ans. a=l—(n−1)¢, S=3n[2l—(n−1)è]. 3d. Given a, n, l to find &, S?............................................Ans. d= S={n(a+1). 2S―n(n−1)♪ 1_2S+n(n−1)♪ ̧ 4th. Given d, n, S to find a, 1?.Ans. a 2n 5th. Given a, n, S to find c, l?................ Ans. d= n-1' 2n n(n−1) +1, s=(?+a)(l—a+ô) 8th. Given a, 1, S to find n, &?..............Ans. d= } Ans. la+(n-1)♪, n=` 10th. Given 8, 1, S to find a, n? Ans.a−b−(n−1), n= Examples: 28 (l+a)(1-a) 2.8 2S−(1+a)' "a+l' -2a+√(-2a)2+8&S 28 8+21+√(8+27)2—8&S 28 ........Ans. 75. ...Ans. —}. 1st. Find the 13th term of the series 3, 9, 15, &c.?....... ..Ans. 363. .....Ans. 16. ......Ans. —841. Ans. 33, 44, 44, 52, 6, 63, 71, 7, 8. 8th. Insert 8 arithmetical means between -3 and -3? Ans. -23, -2, -24, -2, -12, −1, −14, -1. 9th. Insert 7 arithmetical means between -13 and 3? Ans. 11, -9, −7, —5, —3, −1, +1. OF GEOMETRICAL PROGRESSION. 187. If the continued geometrical proportion a: bbc is extended indefinitely, the series of equal ratios forms a geometrical progression, or progression by quotient. In writing the geometrical progression the terms which are consequent and antecedent, in succession, are not repeated as in the continued geometrical proportion, the notation of which a bbc c d :: d: e, &c. is replaced by the following a:b:c:d: e, &c. Both forms of expression have the same meaning. The geometrical progression is hence composed of a series of terms such that, if each term is divided by the term which precedes it, the quotient is a constant quantity; this quotient is the ratio of the progression. Denoting the ratio by q, when q>1 the series is increasing, and when q<1 the series is decreasing. 26 18: 54: 162: 486, &c. 180 60 20:20:20:29, &c. are examples of geometrical progressions in numbers. The ratio of the first, which is an increasing series, is 3, and the ratio of the second, which is a decreasing series, is 188. Leta: b:c: d kl, be the general expression of a geo metrical progression whose ratio is q. Since b=aq and c=bq... c=aq2. d=cq=bq2=aq3: e=dq=aq^. The exponent of q in the first term a, which has 0 term before it, is 0=1-1. The exponent of q in the second term aq or b, which has 1 term before it, is 1=2-1. The exponent of q in the third term aq2 or c, which has 2 terms before it, is 2=3-1. The exponent of q in the fourth term aq3 or d, which has 3 terms before it, is 3=4-1. The exponent of q in the nth term aq-1 or 1, which has n-1 terms before it, is n-1. Whence the exponent of q in any term is indicated by the number of preceding terms, and any term whatever is equal to the first term multiplied by the ratio raised to a power marked by the number of the terms which precede the term proposed. The formula l-aq-1 is called the general term of a geometrical progression. 1 If a and I are the extremes, and n the number of terms of a progression, abc:def:...k: l, whose ratio is q. progression may be Since b=aq, c=aq2...k=aq" 2, l=aq"-1, the same expressed also thus, a: aq: aq2 : aq3. aq"-2: aq"-1. .... If it is required to insert m geometrical means between a and 7, since the progression is to consist of two extremes, a, l, and m mean terms, the whole number of terms is m+2. Substituting m+2 for n, in formula 1, l=aqTM+1 qm + 1 = a a 2. Hence the determination of the constant ratio by means of which the second and following terms are formed from the first term and from each other (or from the general term, formula 1) depends on the extraction of the m+1th root; an operation which, in particular cases, is much facilitated by the use of logarithms. From formula 2 it follows that if the same number of geometrical means is inserted between the 1st and 2d, the 2d and 3d,.... the (n-1)th and nth terms of a geometrical progression, the series composed of these partial progressions is a geometrical progression. 2d term For if is replaced, in formula 2, successively by 1st term 3d term 2d term a nth term =q. ・(n−1)th term '′; since q' is constant, q="+, the ratio ... of all the partial progressions must be constant; and as that term which is the last of any one of the partial progressions is also the first term of the following progression, it is evident that the series composed of these partial progressions is a geometrical progression. |