Thus, if the series be 2, 4, 6, 8, 10, then will 2 + 10 = 4+8=2X 6 = 12. And, if the series be a, a +d, a + 2d, a + 3d, a + 4d, then will a +(a+4d) = (a + d) + (a + 3d) = 2 × (a + 2d). 5. The last term of any increasing arithmetical series is equal to the first term plus the product of the common difference by the number of terms less one; and if the series be decreasing, it will be equal to the first term minus that product. Thus, the nth term of the series a, a +d, a + 2d, a + 3d, a + 4d, &c., is a +(n−1)d. α And the nth term of the series a, a d, a 2d, a 4d, &c., is a — (n − 1)d. 3d, 6. The sum of any series of quantities in arithmetical progression is equal to the sum of the two extremes multiplied by half the number of terms. Thus, the sum of 2, 4, 6, 8, 10, 12, is = (2 + 12) × 42. 6 2 14 X 3 And if the series be a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d), &c. . . . +l, and its sum be denoted by S, we shall have S= (a + 1) X number of terms. η 2' where is the last term, and n the Or, the sum of any increasing arithmetical series may be found, without considering the last term, by adding the product of the common difference by the number of terms less one to twice the first term, and then multiplying the result by half the number of terms. And, if the series be decreasing, the sum will be found by subtracting the above product from twice the first term, and then multiplying the result by half the number of terms, as before. Thus, if the series be a + (a + d) + (a + 2d) + (a + 3d) +(a + 4d), &c., continued to n terms, we shall have And if the series be a + (ad) + (a - 2d) + (a-3d) + (a — 4d), &c., to n terms, we shall have (*) The sum of any number of terms (n) of the series of natural n(n+1) numbers 1, 2, 3, 4, 5, 6, 7, &c., is 2 EXAMPLES. 1. The first term of an increasing arithmetical series is 3, the common difference 2, and the number of terms 20; required the sum of the series. First, 32(201)=3+2 X 193 +38 41, the last term. 20 Or, { 2 × 3+ (20 − 1) × 2}=(6+19×2)×10= (6 + 38) × 10 = 44 × 10 = 440, as before. 2. The first term of a decreasing arithmetical series is 100, the common difference 3, and the number of terms 34; required the sum of the series. = First, 100-3(34 — 1) — 100 — 3 × 33 — 100 — 99 — 1, the last term. 17 = (200 (200-99) 99) × 17 = 101 × 17 = 1717, as before. 3. Required the sun of the natural numbers 1, 2, 3, 4, 5, 6, &c., continued to 1000 terms. Ans. 500500. 4. Required the sum of the odd numbers 1, 3, 5, 7, 9, &c., continued to 101 terms. Ans. 10201. 5. How many strokes do the clocks of Venice, which go on to 24 o'clock, strike in a day? Ans. 300. 6. Required the 365th term of the series of even numbers 2, 4, 6, 8, 10, 12, &c Ans. 730. 7. The first term of a decreasing arithmetical series is 10, Thus, 1+2+3+4+5, &c., continued to 100 terms, is 50 X 1015050. 100×101 2 Also the sun of any number of terms (n) of the series of odd numbers 1, 3, 5, 7, 9, 11, &c., is = n2. Thus, 14-3+5+7+9, &c., continued to 50 terms, is 502-2500. And if any three of the quantities, a, d, n, S, be given, the fourth may be found from the equation a ± Where the upper sign is to be used when the series is increasing, and the lower sign when it is decreasing; also the last term 7 = (n-1)d, as above. the common difference 1 3' and the number of terms 21; re quired the sum of the series. Ans. 140. 8. One hundred stones being placed on the ground, in a straight line, at the distance of a yard from each other, how far will a person travel, who shall bring them one by one, to a basket, placed at the distance of a yard from the first stone? Ans. 5 miles and 1300 yards. OF GEOMETRICAL PROPORTION AND PROGRESSION. * GEOMETRICAL PROPORTION, is the relation which two quantities of the same kind have to two others, when the * If there be taken any four proportionals, a, b, c, d, which it has been usual to express by means of points: thus, a: bc:d, a C b d where the equal this relation will be denoted by the equation ō ratios are represented by fractions, the numerators of which are the antecedents, and the denominators the consequents. Hence, if each of the two members of this equation be multiplied by bd, there will arise ad=bc. From which it appears, as in the common rule, that the product of the two extremes of any four proportionals is equal to that of the means. And if the third c, in this case, be the same as the second, or C b b, the proportion is said to be continued, and we have ad=b2, or vad; where it is evident, that the product of the extremes of three proportionals is equal to the square of the mean; or, that the mean is equal to the square root of the product of the two extremes. Also, if each member of the equation ad=bc be successively divided by bd, dc, ae, &c., the results will give So that, by following this method, we can easily obtain all the transformations of the terms of the proportion, that can be made to agree with the equations ad=bc. In like manner, from the same equality a C b d' there will result, by ma nc ma mc multiplication, the following equivalent forms: mb Which, being converted into proportions, become ma mb :: ne: nā, and ma: nbmc: nd. And, by taking any like powers, or roots, of the antecedents, or leading terms of each pair, are the same parts of their consequents, or the consequents of the antecedents. In which cases terms in the form of a proportion, am : bn :: cm : dr. aXeXi,&c. cXgXl,&c. bXfXk,&c. d×h×m,&c. Or, by putting the expression in the form of a proportion, aci, &c., nd ; which, being expressed in the form of a proportion, gives ma±nb: nb :: mcnd: nd; or ma±nb: mc±nd :: nb: nd. And if the abovementioned equation a C be put, by a similar mul b tiplication of its terms, under the form pa pc and then augmented or qb diminished by 1, as in the last case, there will arise pa±qb: pc±qd :: qb: qd. Whence, dividing each of the antecedents of these two ana Or, by converting the corresponding terms of this equation into a proportion manb: mc±nd::pa ±qb: pc qd. Also, because the common equation lent forms we shall obtain, by a similar process, ma±nc: pa±qc :: mb±nd: pb± qd; which two analogies may be considered as general formulæ for changing the term sof the proportion a:b::c:d, without altering its nature. Thus, by supposing m, n, p, q, to be each = 1, and taking the antecedents with the superior signs, and And if two quantities only are to be compared together, the part or parts, which the antecedent is of its consequent, or the consequent of the antecedent, is called the ratio; observing, in both cases, always to follow the same method. Hence, three quantities are said to be in geometrical proportion, when the first is to the same part, or multiple, of the second, as the second is of the third. Thus, 3, 6, 12, and a, ar, ar2, are quantities in geometrical proportion. And four quantities are said to be in geometrical proportion, when the first is the same part, or multiple, of the second, as the third is of the fourth. Thus, 2, 8, 3, 12, and a, ar, b, br, are geometrical proportionals. Direct proportion, is when the same relation subsists between the first of four terms and the second, as between the third and fourth. Thus, 3, 6, 5, 10, and a, ar, b, br, are in direct proportion. Inverse, or reciprocal proportion, is when the first and second of four quantities are directly proportional to the reciprocals of the third and fourth. the consequents with the inferior, we have a+b: a-b::c+d: c- d, and aca-c::b+d:b−d; which forms, together with several of those already given, are the usual transformations of the common analogy pointed out above. In like manner, by taking m, n, and p each = 1, and q=0, there will arise ab: a::cd: c, and a±c:a :: b±d:b; each of which proportions may be verified by making the product of the extremes equal to that of the means, and observing that ad bc. Lastly, taking any number of equations of the form before used, for expressing proportions, to a b се g d f &c.; which, according to the common method, are called a series of equal ratios, and are usually denoted by a b c d e f :: g: h::, &c., we shall necessarily have, from the fractions being all equal to each other, e f g a C b d q, =q, &c. And, by multiplying q by each of the denominators, h a = bq, c=dq, e = fq, g = hq, &c. Whence, equating the sum of all the terms on the lefthand side of these equations, with those on the right, we have a+c+e+g+,&c. =(b+d+ f+h+, &c.,)q. And consequently, by division, and the properties of proportionals before shown, a+c+e+g+, &c. b+d+f+h+,&c. b+d b+d+ƒ ̃ which results show that, in a series of equal ratios, the sum of any number of the antecedents is to that of their consequents, as one, or more of the antecedents, is to one, or the same number of consequents. |