531. EXAMPLES. 1. The first term of an arithmetical series is 2, the last 364, the number of terms 12; what is the sum of the series? Answer, 2196. 2. What is the sum of a decreasing arithmetical series whose first term is 102, last term 8, and number of terms 8 ? Answer, 376. RULE 2. n - To find the common difference of an arithmetical series. 532. Divide the difference between the first and last term by the number of terms less one; d, ord. That this rule is necessarily correct, the student has but to consider that the last term of every arithmetical series is formed by adding to the first term as many times the common difference, as the number of terms less one; see paragraphs 525-6. Thereforeƒ+ (n − 1) d = l, and by transposition (n − 1) d = l-f f, consequently d = N- 1. EXAMPLES. 1. The last term of an arithmetical series is 364, the first term 2, and the number of terms 12; what is the common difference ? Answer, 321. 2. What is the common difference of a descending arithmetical series, the first term of which is 102, last term the number of terms 8? RULE 3. 8, and Answer, 15. To find the number of terms of an arithmetical series, the first and last term, and common difference, being given. 534. Divide the difference between the first and last terms l-f by the common difference, and add I to the quotient; 1. The principle of this rule may be thus explained.— The last term is formed by adding the first, and as many times the common difference as the number of terms less one; therefore by paragraph 521, ƒ + (n − 1) d = 7; then (n − 1) d=1-f; and n — 1= f and n= 531. EXAMPLES. +1. 1. The first term of an arithmetical series being 2, the last term 364, and the common difference 321; what is the num ber of terms? Answer, 12, 2. What is the number of terms of a decreasing arithmetical series, of which the first term is 102, the last term — 8, and the common difference 15? Answer, 8. RULE 4. To find the last term, or any intermediate term between the first and last of an arithmetical series, the first term, the number of terms, and the common difference, being given. 536. Multiply the number of the terms, diminished by unity, by the common difference, to the product of which add the first term, if the series be an increasing one; but take the dif ference between the product and the first term if the series be a decreasing one. 537. For the principle on which this rule is founded, see paragraph 521; ƒ + (n − 1) d = l, so also ƒ + d = second term, f2d third term, ƒ+ 12d = thirteenth term, &c. 538. EXAMPLES. 1. The first term of an arithmetical series being 2, the number of terms 12, and the common difference 321; what is the last term? Answer, 364. 2. What are the 3rd, 6th, 8th, and 9th terms of an arithmetical series, the first term being 2, and the common difference 321, as above? Answers, 67, 166, 2324, and 265. 3. What is the last term of a decreasing arithmetical series, the first term of which is 102, the number of terms 8, and the common difference 15 ? RULE 5. Answer, 8. To find the first term of an arithmetical series, the last term, the number of terms, and the common difference, being given. 539. Multiply the number of terms less one by the common difference, and subtract the product from the last term, if the series be an increasing one; but add the product to the last term, if it be a decreasing series, viz., l - (n − 1) d = f. As every arithmetical series is formed by the continual addition or subtraction of some particular number, as a, a + d, a + 2d, a+3d, a + 4d, we easily see, that if from the last term we subtract the product of the common difference, multiplied by the number of the terms diminished by unity, the remainder will be the first term. 540. EXAMPLES. 1. The last term of an arithmetical series being 364, the number of terms 12, and the common difference 32; what is the first term? Answer, 2. 2. What is the first term of an arithmetical series, the last - 8, the number of terms 8, and the common Answer, 102. term of which is difference, 15 ? 541. PROMISCUOUS EXAMPLES. 1. A person owing a certain sum of money, paid it in thirteen separate payments, increasing in arithmetical progression, the first payment was 3s., and the last was £2 11's. How much did each payment exceed the former, and how much did he owe in all ? Ans., Diff., 4s. ; debt, £17 11s. 2. A person travelling from one town to another, went six miles the first day, nine miles the second, and so on in arithmetical progression until his last day's journey was sixty miles. Required the distance of the towns from each other, and how long he was on his journey? Ans., 19 days; 627 miles. 3. A sum of money is divided among twelve persons whose shares are in arithmetical progression, the first received 3s., and the second 7s., and so on. Required the sum divided, and the share of the last? Ans., £15 divided; last person's share, £2 9s. 4. If fifty eggs be placed in a right line, exactly a yard asunder, and the first a yard from a basket, what length of ground must a person go who, going from the basket, gathers them singly, and returns with them one by one to the basket? Ans., 2500 yds. 5. If twenty oranges be placed in a right line, exactly two yards asunder, and the first one yard from a basket, how many yards will a person walk who, beginning at the basket, gathers them one by one into the basket? Ans., 2780 yds. GEOMETRICAL PROGRESSION. 542. A series of numbers increasing or decreasing regularly, by the continual multiplication or division of some particular number, is said to be in geometrical proportion. Such a set of numbers is called a geometrical series. Thus, 3, 6, 12, 24, 48, 96, &c., are numbers in geometrical proportion. In this series each antecedent* is to its consequent+ as 1 is to 2: each consequent being twice its antecedent. The number by which each term of a geometrical series is multiplied to produce the next, is called the common ratio. This common ratio may always be found by dividing any term in the series by the one before it, so that in the above series the common ratio is 2. When * Antecedent, a thing going before another, either in order of time or place; from ante, before, and ced-o, to go. + Consequent, that which comes after or follows another; from con, with or together, and sequ-or, to follow. the common ratio is larger than unity, the terms of the series regularly increase, as in the above series. When the common ratio is less than unity, the terms of the series regularly decrease, as in the series 32, 16, 8, 4, 2, 1, 4, 4, &c., in which the common ratio is . When the terms of a series increase, such series is called an ascending series. When the terms of the series decrease, such series is called a descending series. 543. In geometrical progression five things are to be considered, viz. : 1. The first term, which may be designated by f 3. The number of terms, 4. The common ratio, 5. The sum of the terms If any three of the above be given, the others are easily found. 541. All the terms of a geometrical series comprehended between the first and last term, are called mean proportionals, or geometrical means. Thus, in the series 3, 12, 48, 192, the twe mean proportionals between 3 and 192 are 12 and 48. 545. In every geometrical series, composed of an even number of terms, the product of the extremes is equal to the product of the two means, or of any two mean proportionals equally distant from the extremes. Thus, in the series 2, 4, 8, 16, 32, 64, the product of the extremes 2 and 64 is equal to the product of the means 8 and 16, and also to the product of the mean proportionals 4 and 32; for 64 × 2 = 32 × 4 = X 8 = 128, = 16 546. In every geometrical series composed of an odd number of terms, the square of the middle term is equal to the product of the extremes, or to the product of any two mean proportionals equally distant from the middle term. Thus, in the series 2, 4, 8, 16, 32, the square of 8, which is 64, is equal to the product of 32 by 2, or of 16 by 4; for 8 × 8=32 × 2 = 16 × 4 ; hence, the geometrical mean between any two terms of a series is equal to the square root of their product. 547. In every geometrical series, each antecedent is to its consequent as the first term is to the second, or as unity is to the ratio of the progression. 548. When we know the first term and the ratio of a geometrical series, we can find any other term; the first term multiplied by the common ratio becomes the second term, the second multiplied by the common ratio becomes the third, and so on progressively. Thus, in series 1,3,9, 27, &c. the common ratio is 3, and the first term multiplied by 3 becomes the second term, and generally any term in the series multiplied by 3 becomes the subsequent term. In this manner may we find any term in a geometrical series, the first term and common ratio of which is known; but when a series is composed of many terms, to find the last, or any term at a considerable distance from the first term, by this continued multiplication of each successive term by the common ratio would be exceedingly tedious; the object, however, is easily and quickly obtained by having recourse to the powers of numbers and their exponents. Before we proceed further, perhaps it may be well for the student carefully again to read over Chapter 16, which treats of powers and their exponents. 549. If we take the series above, 1, 3, 9, 27, &c., the common ratio of which is 3, we perceive that the first term multiplied by the common ratio becomes the second term; that the first term multiplied by the square or second power of the ratio becomes the third term; that the first term multiplied by the cube, or third power of the ratio, becomes the fourth term, and so on. So that if the first term be represented by f, and the ratio by r, the general expression for any geometrical series will be, f, fr, fr2, fr3, fr4, &c. Here are five terms, the fifth and last is the product of the first term by the 4th power of the ratio; the fourth term is the product of the first term by the 3rd power of the ratio, and so on; hence we perceive that any term of a geometrical series may be found by multiplying the first term by such a power of the ratio, as is indicated by the number of terms less 1. Thus, if ƒ 3, and r = 4; then 3 × (4 × 4 X 4) = 3 × 64 = 192 4th term. That this result is correct may be show by carrying on the series to the 4th term, viz., 3, 12, 48, 192. From this it will be seen, that if the first term of a geometrical series be unity, the succeeding terms will be the progressive powers of the ratio; as Exponents of powers, Powers of the ratio, and 0 1 2 3 4 5 terms of the series, 1, 4, 16, 6 64, 256, 1024, 4096 Exponents of powers, Powers of the ratio, Terms of the series, Number of the terms, 551. The above table exhibits a series, the first term of which is 8, the ratio 3, and the number of terms 6. Now, we may find the last, or sixth term of this series in three ways, without producing the intermediate terms. First. By multiplying the first term by the fifth power of the ratio, as 243 X 550. That if the first term of the series be any other number than unity, the value of any term will correspond with the product of the first term, by such a power of the ratio as shall correspond with the number of the terms sought less 1 ; as 0 1 2 3 4 5 1 3 9 27 81 243 8, 24, 72, 216, 648, 1944 1 2 3 4 5 6 |