ARITHMETICAL PROGRESSION. Any rank, or series of numbers, consisting of more than two terms, which increases or decreases by a common difference, is called an Arithmetical series, or progression. When the series increases, that is, when it is formed by the constant addition of the common difference, it is called an ascending series, thus, 1, 3, 5, 7, 9, 11, &c. Here it will be seen that the series is formed by a continual addition of 2 to each succeeding figure. When the series decreases, that is, when it is formed by the constant subtraction of the common difference, it is called a descending series, thus, 14, 12, 10, 8, 6, 4, &c. Here the series is formed by a continual subtraction of 2, from each preceding figure. The figures that make up the series are called the terms of the series. The first and last terms are called the extremes, and the other terms, the means. From the above, it may be seen, that any term in a series may be found by continued addition or subtraction, but in a long series this process would be tedious. A much more expeditious method may be found. 1. The ages of six persons are in arithmetical progression. The youngest is 8 years old, and the common dif. ference is 3, what is the age of the eldest? In other words, what is the last term of an arithmetical series, whose first term is 8, the number of terms 6, and the common differ. ence 3? 8, 11, 14, 17, 20, 23. Examining this series, we find that the common difference, 3, is added 5 times, that is one less than the number of terms, and the last term, 23, is larger than the first term, by five times the addition of the common difference, three; Hence the age of the elder person is 8+3×5=23. Therefore when the first term, the number of terms, and the common difference, are given, to find the last term, Multiply the common difference into the number of terms, less 1, and add the product to the first term. When are numbers said to be in arithmetical progression? 2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term? Ans. 301. 3. There are, in a certain triangular field, 41 rows of corn; the first row, in 1 corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row? A. 81 hills. 4. A man puts out $1 at 6 per cent. simple interest, which, in 1 year, amounts to $1,06 in 2 years to $1,12, and so on, in arithmetical progression, with a common difference of $0,06; what would be the amount in 40 years? A. $3,40. Hence we see, that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the principal is the first term, the last amount is the last term, the yearly interest is the common difference, and the num ber of years is 1 less than the number of terms. It is often necessary to find the sum of all the terms, in an arithmetical progression. The most natural mode of obtaining the amount would be to add them together, but an easier method may be discovered, by attending to the following explanation. 1. Suppose we are required to find the sum of all the terms, in a series, whose first term is 2, the number of terms 10, and the common difference 2. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 20, 18, 16, 14, 12, 10, 8, 6, 4, 2 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, The first row of figures above, represents the given series. The second, the same series with the order inverted, and the third, the sums of the additions of the corresponding terms in the two series. Examining these series, we shall find that the sums of the corresponding terms are the same, and that each of them is equal to the sum of the extremes, viz. 22. Now as there are 10 of these pairs in the two series, the sum of the terms in both, must be 22× 10=220. But it is evident, that the sum of the terms in one series, can be only half as great as the sum of both, therefore, if we divide 220 by 2, we shall find the sum of the terms in one series, which was the thing required. 220—2—110, the sum of the given series. From this illustration we derive the following rule; When the extremes and number of terms are given, to find the sum of the terms, Multiply the sum of the extremes by the number of terms, and divide the product by 2. 2. The first term of a series is 1, the last term 29, and the number of terms 14. What is the sum of the series? A. 210. 3. 1st term, 2, last term, 51, number of terms, 18. Required the sum of the series. A. 477. 4. Find the sum of the natural terms 1, 2, 3, &c. to 10,000. A. 50,005,000. 5. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money ? The last year's rent will evidently be $50 without interest, the last but one will be the amount of $50 for 1 year, the last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years= $107. = If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series? A. 1570. 6. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent. simple interest? A. $7900. 7. There are, in a certain triangular field, 41 rows of corn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how many hills of corn in the field? A. 1681. The method of finding the common difference, may be learned by what follows. 1. A man bought 100 yards of cloth in Arithmetical progression for the first yard he gave 4 cents, and for the last 301 cents, what is the common increase on the price of each yard? As he bought 100 yards, and at an increased price upon every yard, it is evident that this increase was made 99 times, or once less than the number of terms in the series. Hence the price of the last yard was greater than the first, by the addition of 99 times the regular increase. Therefore if the first price be subtracted from the last, and the remainder be divided by the number of additions (99), the quotient will be the common increase; 301—4= 297 and 297-99=3, the common difference. Hence, when the extremes and number of terms are given, to find the common difference, Divide the difference of the extremes, by the number of terms less 1. 2. Extremes 3 and 19; number of terms 9. the common difference. Required 3. Extremes 4 and 56; number of terms 14. Required the common difference. A. 4. 4. A man had 15 houses, increasing equally in value, from the first, worth $700, to the 15th, worth $3500. What was the difference in value between the first and second? A. 200. In Arithmetical progression, any three of the following terms being given, the other two may be found. 1. The first term. 2. The last term. 4. The common difference. 3. The number of terms. 5. The sum of all the terms. GEOMETRICAL PROGRESSION. Any series of numbers, consisting of more than two terms, which increases by a common multiplier, or decrea ses by a common divisor, is called a Geometrical Series. Thus the series 2, 4, 8, 16, 32, &c. consists of terms, each of which is twice the preceding, and this is an increasing or ascending Geometrical series. The series 32, 16, 8, 4, 2, consists of numbers, each of which is one half the preceding, and this is a decreasing or descending Geometrical series. The common multiplier or divisor is called the Ratio, and the numbers which form the series are called Terms. As in Arithmetical, so in Geometrical progression, if In arithmetical progression, what are the terms used, and what are the rules for finding them? any three of the five following terms be given, the other two may be found. 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms. 1. A man bought a piece of cloth containing 12 yards, the first yard cost 3 cents, the second 6, the third 12, and so on, doubling the price to the last, what cost the last yard? 8×2×2×2×2×2×2×2×2×2×2×2=3x211=6144 Ans. In examining the above process, it will be seen, that the price of the second yard is found by multiplying the first payment into the ratio (2) once; the price of the third yard, by multiplying by 2 twice, &c., and that the ratio (2) is used as a factor eleven times, or once less than the num. ber of terms. The last term then, is the eleventh power of the ratio (2) multiplied by the first term (3). Hence the first term, ratio, and number of terms, being given, to find the last term. Multiply the first term, by that power of the ratio, whose index is one less than the number of terms. Note. In involving the ratio, it is not always necessary to produce all the intermediate powers; the process may often be abridged, by multiplying together two powers already obtained, thus, The 11th power — =the 6th power X the 5th power, &c. 2. If the first term is 2, the ratio 2, and the number of terms 13, what is the last term? A. 8,192. 3. Find the 12th term of a series, whose first term is 3, and ratio, 3. A. 531,441. 4. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels; these he plants the second year; now, supposing the annual increase to continue 8 fold, what would be the produce of the 16th year, allowing 1000 kernels to a pint? A. 2199023255.552 bushels. 5. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years? 217131072. A. 1310.72. When are numbers in geometrical progression? What are the terms used? What are the rules for finding them? |