OF ARITHMETICAL PROPORTION AND PROGRESSION. ARITHMETICAL PROPORTION, is the relation which two quantities of the same kind, have to two others, when the difference of the first pair is equal to that of the second. Hence, three quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the second and third. Thus, 2, 4, 6, and a, a+b, a+26, are quantities in arithmetical proportion. And four quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the third and fourth. Thus, 3, 7, 12, 16, and a, a+b, c, c+b, are quantities in arithmetical proportion. ARITHMETICAL PROGRESSION is when a series of quantities increase or decrease by the same common differ ence. Thus, 1, 3, 5, 7, 9, &c. and a, a+d, a+2d, a+3d, &c. are increasing series in arithmetical progression, the common differences of which are 2 and d. -- And 15, 12, 9, 6, &c. and a, a — d, a — 2d, a—3d, &c. are decreasing series in arithmetical progression, the common differences of which are 3 and d. The most useful properties of arithmetical proportion and progression are contained in the following theorems : 1. If four quantities are in arithmetical proportion, the sum of the two extremes will be equal to the sum of the two means. Thus if the proportionals be 2, 5, 7, 10, or a, b, c, d; then will 2+10=5+7, and a+d=b+c. 2. And if three quantities be in arithmetical propor tion, the sum of the two extremes will be double the mean. Thus, if the proportionals be 3, 6, 9, or a, b, c, then will 3+9=2x6=12, and a+c=2b. 3. Hence an arithmetical mean between any two quantities is equal to half the sum of those quantities. Thus, an arithmetical mean between 2 and 4 is 2+4 4. In any continued arithmetical progression, the sum of the two extrenies is equal to the sum of any two terms that are equally distant from them, or to double the middle term, when the number of terms is odd Thus, if the series be 2, 4, 6, 8, 10, then will 2+10= 4+8=2×6=12. And, if the series be a, a+d, a+2d. a+3d, a+4d, then will a+(a+4d)=(a+d)+a+ d) = x(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. * If two, or more, arithmetical means between any two quantities be required, they may be expressed as below. 2a+b a+26 3 and - two arithmetical means between a and b, a 3 Thus, being the less extreme and b the greater. na+b (n-1)a+2b (n−2)a + 3b a+nb And &c. to any number (n) b-a of arithmetical means between a and b; where the common differ n+1 ence; which being added to a, gives the first of these means; and the again to this last, gives the second; and so on. And the nth term of the series a, a―d, a-2d, a-3d, a-4d, &c. is a-(n−1)d. 6. The sum of any series of quantities in arithmetica! progression is equal to the sum of the two extremes multiplied by half the number of terms. 6 Thus, the sum of 2, 4, 6, 8, 10, 12, is = =14X3=42. (2+12) × And if the series be a+(a+d)+(a+2d)+(a+3d)+ (a+4d)&c. +1, and its sum be denoted by S, we shall have S=(a+1)×2, where I is the last term, and ʼn the number of terms. 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, its 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 S= {2a+(n-1)d} ×2. And if the series be a+(a-d)+(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 numbers 1, 2, 3, 4, 5, 6, 7, &c. is = n(n+1) Thus, 1+2+3+4+5, &c. continued to 100 terms, is X 101 == 5050. 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, 3+2(20-1)=3+2×19=3+38=41, the last term. 20 20 2 And (3+41)x44x44x10=440, the sum required. 20 Or, {2×3+(20—1)×2}x=(6+19×2) X10=(6 +38) X10=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)=100-3X33-100-99=1, the last term. 34 Or, {2×100—(34—1) ×S} ×34=(200 – 33×3) × 17 2 =(200-99) X17=101X17=1717, as before. 3. Required the sum of the natural numbers, 1, 2, 3, 4, 5, 6, &c. continued to 1000 terms. 4. Required the sum of the odd numbers &c. continued to 101 terms. Ans. 500500. 1, 3, 5, 7, 9, Ans. 10201. Also the sum of any number of terms (n) of the series of odd numbers 1, 3, 5, 7, 9, 11, &c. is = n2. Thus, 1+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 s= {2a±(n−1)dx, or (a+1)x12 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 = a+ (n-1)d, as above. 5. How many strokes do the clocks of Venice, which Ans. 300. go on to 24 o'clock, strike in a day? 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, the common difference and the number of terms 1 3' 21; required 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: b::c:d, this relation will be denoted by the equation α C ;; where the equal 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 cb, the proportion is said to be continued, and we have ad=b2, or bad; 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. |