THEOREM 3. The difference between the extreme terms of an arithmetical progression, is equal to the common difference of the series multiplied by one less than the number of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the common difference is 2, and one less than the number of terms 9; then the difference of the extremes is 20 - 2 = 18, and 2 x 9 = 18 also. Consequently the greatest term is equal to the least term added to the product of the common difference multiplied by one less than the number of terms. Theorem 4. The sum of all the terms of any arithmetical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided hy 2; or the sum of the two extremes multiplied by the number of the terms, gives twice the sum of all the terms in the series. This is made evident by setting the terms of the series in an inverted order, under the same series in a direct order, and adding the corresponding terms together in that order. Thus, in the series 1, 3, 5, 7, 9, 11, 9, 7, 5, 3, the sums are 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16, which must be double the sum of the single series, and is equal to the sum of the extremes repeated as often as are the number of the terms. From these theorems may readily be found any one of these five particulars ; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given, as in the following problems. 11, 13, 15; 13, PROBLEM I. Given the extremes, and the number of terms, to find the sum of all the terms. Add the extremes together, multiply the sum by the number of terms, and divide by 2. EXAMPLES. 2 1. The extremes being 3 and 19, and the number of terms 9; required the sum of the terms. 19 + 3 22 Here X 9 x 9= 11 x 9 = 99, the answer. 2 2. It is required to find the number of all the strokes a common clock strikes in one whole revolution of the index, or in 12 hours. Ans. 78. 3. How many strokes do the clocks of Venice strike in the compass of a day, which go continually on from 1 to 24 o'clock ? Ans. 300. 4. What debt can be discharged in a year, by weekly payments, in arithmetical progression, the first payment being 1s, and the last or 52d payment 51 3s ? Ans. 1351 4s. PROBLEM II. Given the extremes, and the number of terms, to find the common difference. Subtract the less extreme from the greater, and divide the remainder by 1 less than the number of terms, for the common difference. EXAMPLES. 1. The extremes being 3 and 19, and the number of terms 9; required the common difference. 19-3 16 Here 2, Answer. 2. If the extremes be 10 and 70, and the number of terms 21 ; what is the common difference, and the sum of the series ? Ans., the common difference is 3, and the sum is 840. 3. A certain debt can be discharged in one year, by weekly payments in arithmetical progression, the first payment being ls, and the last 5l 3s; what is the common difference of the terms ? Ans. 2. PROBLEM III, Given one of the extremes, the common difference, and the number of terms; to find the other extreme, and the sum of the series. MULTIPLY the common difference by 1 less than the number of terms, and the product will be the difference of the extremes: then add the product to the less extreme to give the greater extreme, or subtract it from the greater, to give the less. EXAMPLES. 198 2 = 1. Given the least term 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series. Here 2 x (9—1) + 3 = 19 = the greatest term : hence (19+3) ; 99 = the sum of the series. 2. If the greatest term be 70, the common difference 3, and the number of terms 21, what is the least term, and the sum of the series ? Ans., the least term is 10, and the sum 840. 3. A debt can be discharged in a year, by paying 1 shilling the first week, 3 shillings the second, and so on, always 2 shillings more every week; what is the debt, and what will the last payment be? Ans., the last payment will be 51 3s, and the debt is 1351 4s. PROBLEM IV. To find an arithmetical mean proportional between two given terms. Add the two given extremes or terms together, and take half their sum for the arithmetical mean required. EXAMPLE. Here =9= the mean required. PROBLEM V. To find two arithmetical means between two given extremes. Subtract the less extreme from the greater, and divide the difference by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, will give the means. EXAMPLE. To find two arithmetical means between 2 and 8. -2 3 PROBLEM VI. To find any number of arithmetical means between two given terms or extremes. SUBTRACT the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required. EXAMPLE. To find five arithmetical means between 2 and 14. Here 14 52 = 2 = common difference. 6 Then by adding this common difference continually, the means are found to be 4, 6, 8, 10, 12. See more of arithmetical progression in the Algebra. GEOMETRICAL PROPORTION AND PROGRESSION. * If there be taken two ratios, as those of 6 to 3, and 14 to 7, which, by what has been already said, (p. 50, 75,) may be expressed fractionally, j and ""; to judge whether they are equal or unequal, we must reduce them to a common denominator, and we shall have 6 x 7, and 14 x 3 for the two numerators. If these are equal, the fractions or ratios are equal. Therefore, THEOREM 1. If four quantities be in geometrical proportion, the product of the two extremes will be equal to the product of the two means. And hence, if the product of the two means be divided by one of the extremes, the quotient will give the other extreme. Thus, of the above numbers, if the product of the means 42 be divided by 6, the quotient 7 is the other extreme ; and if 42 be divided by 7, the quotient 6 is the first extreme. This is the foundation of the practice in the Rule of Three. We see, also, that if we have four numbers, 6, 3, 14, 7, such, that the products of the means and of the extremes are equal, we may hence infer the equality of the ratios şi 4, or the existence of the proportion 6 : 3 :: 14 : 7. Hence Theorem 11. We may always form a proportion of the factors of two equal products. If the two means are equal, as in the terms 3, 6, 6, 12, their product becomes a square. Hence. THEOREM 111. The mean proportional between two numbers is the square root of their product. We may without destroying the accuracy of a proportion, give to its various terms all the changes which do not affect the equality of the products of the means and extremes. 14 * See the note at p. 75. = 3 = 3 a 3 Thus, with respect to the proportion 6:3 :: 14: 7, which gives 6 x7=3 x 14, we may displace the extremes, or the means, an operation which is denoted by the word Alternando. This will give 6 : 14 :: 3:7 or 7 : 3 :: 14 : 6 or 7 : 14 :: 3 : 6. Or, 2dly, we may put the extremes in the places of the means, called Invertendo. Thus 3 : 6 :: 7 : 14. Or, 3dly, we may multiply or divide the two antecedents, or the two consequents, by the same number, when proportionality will subsist. As 6 X 4 : 3 :: 14 X 4 :7; viz. 24 : 3 :: 56 : 7 and 6 = 2 : 3 :: 14 • 2:7; viz. 3: 3:: 7:7. Also, applying the proposition in note 2, Addition of Vulgar Fractions, to the terms of a proportion, such as 30 : 6 :: 15 ; 3, or 3 = ', we shall have 30 + 15 15 30 30 + 15 15 and Hence 6 + 3 6 + 3 6 3 Theorem iv. The eum or the difference of the antecedents, is to that of the consequents, as any one of the antecedents is to its consequent. THEOREM v. The sum of the antecedents is to their difference, as the sum of the consequents is to their difference. In like manner, if there be a series of equal ratios, = = 44 = 1; we have Theorem vi. In any series of equal ratios, the sum of the antecedents is to that of the consequents, as any one antecedent is to its consequent. 6 + 10 + 14 + 30 6 10 14 30 Therefore, 3 + 5 + 7 + 15 5 7 15 THEOREM VII. If two proportions are multiplied, term by term, the products will constitute a proportion. Thus, if 30 : 15 :: 6:3 and 2 : 3 :: 4 : 6. or 60 : 45 :: 24 : 18; or Theorem viii. If four quantities are in proportion, their squares, cubes, and all other powers will be in proportion. For this will evidently be nothing else than assuming the proportionality of the products, term by term, of two, three, or more identical proportions. The same properties bold with regard to surd or irrational expressions. Thus, 720 : N 80 :: N 567 : 763 and 12 : N3 :: N 4 in 1. ✓ 720 ✓(9 X 80) 567 For ✓ (9 x 63) and ✓ 63 63 1 2 N 4 = ✓ 3 THEOREM Ix. The quotient of the extreme terms of a geometrical progression is equal to the common ratio of the series raised to the power denoted by 1 less than the number of the terms. So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common ratio is 2, one less than the number of terms 9; then the quotient of the extremes is = 512, and 29 = 512 also. Consequently the greatest term is equal to the less term multiplied by the said power of the ratio, whose index is 1 less than the number of terms. THEOREM X. The sum of all the terms, of any geometrical progression, is 3 3 1 1024 found by adding the greatest term to the difference of the extremes divided by 1 less than the ratio. So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, (whose ratio is 2,) is 1024 - 2 1024 + = 1024 + 1022 = 2046. 2 This subject will be resumed in the algebraic part of this work. A few examples may here be added. . 1 EXAMPLES. 1. The least of ten terms in geometrical progression being 1, and the ratio 2; what is the greatest term, and the sum of all the terms ? Ans., the greatest term is 512, and the sum 1023. 2. What debt may be discharged in a year, or 12 months, by paying 11 the first month, 2l the second, 41 the third, and so on, each succeeding payment being double the last; and what will the last payment be? Ans., the debt 40951 and the last payment 20481. PROBLEM I. To find one geometrical mean proportional between any two numbers. MULTIPLY the two numbers together, and extract the square root of the product, which will give the mean proportional sought. EXAMPLE. To find a geometrical mean between the two numbers 3 and 12. (12 x 3) = N 36 = 6 = the mean. PROBLEM II. To find two geometrical mean proportionals between any two numbers. Divide the greater number by the less, and extract the cube root of the quotient, which will give the common ratio of the terms. Then multiply the least given term by the ratio for the first mean, and this mean again by the ratio for the second mean; or, divide the greater of the two given terms by the ratio for the greater mean, and divide this again by the ratio for the less mean. EXAMPLE. To find two geometrical means between 3 and 24. PROBLEM III. To find any number of geometrical means between two numbers. Divide the greater number by the less, and extract such root of the quotient whose index is 1 more than the number of means required; that is, the 2d root for one mean, the 3rd root for two means, the fourth root for three means, and so on; and that root will be the common ratio of all the terms: then, with the ratio, multiply continually from the first term, or divide continually from the last or greatest term. |