1st. The first term. 3d. The number of terms. 2d. The last term. 4th. The common difference. 5th. The sum of all the terms. Any three of which being given, the other two may be found. CASE I-The first term, common difference, and number of terms given, to find the last term. RULE.-Multiply the number of terms, less 1, by the common difference, and add the first term to the product, the sum will then be the last term, or answer sought. EXAMPLES. 1. A merchant sold 200 yards of cloth; for the first yard he received 4 cents, 7 cents for the second, 10 cents for the third, and so on, with a common difference of three cepts; what did he receive for the last yard? Ans. 601 199 the number of terms, less 1. 597 4 the first term added. cents=$6,01 cent. DEM.-It is plain, that each term ex-. ceeds that preceding it by the common difference; it is then evident, that Ans. 601 cents, he received for the last yard. the last term exceeds the first by as many times the common difference, as there are terms after the first; therefore the last term must equal the first, and the number of terms, less one, repeated by the common difference, as our example plainly shows. 2. A man put out $100, at 7 per cent, simple interest, which amounted to $107 in a year, $114 in 2 years, and so on, in arithmetical progression, with the common difference of $7; what was the amount due at the expiration of 50 years? Ans. $450. 3. John owes William a certain sum, to be paid in arithmetical progression; the first payment is 6 pence, the number of payments 52, and the common difference of the payments is 12 pence; what is the last payment? Ans. £2 11s. 6d. CASE II-The first term, common difference, and the number of terms given, to find the sum of all the terms. RULE.-Multiply half the sum of the extremes by the number of terms, and the product will be the answer or sum of all the terms, he EXAMPLES. 1. A person bought 16 yards of cloth; for the first yard gave 5 pence, for the second 9 pence, and so on, in arithmetical progression; what did the last yard cost him, and what did he pay for the whole? [preceding case. Ans. 65 pence the cost of the last yard, obtained by the 35 half the sum of the extremes. 210 35 12)560 the sum of all the terms. 20)46 8d. A. £2 6s. 8d. the cost of 16 yards. DEM. The price of each succeeding yard being increased by a constant excess, it is plain, that the average price is as much less than the price of the last yard, as it is greater than the price of the first yard; hence it is evident, that half the sum of the first and last, is the average price, and it is also obvious, that the average price multiplied by the whole number of yards, that is, terms, gives the price of the whole number of yards. 2. A man bought 17 yards of Irish linen; for the first yard he gave 2 shillings, for the last 10 shillings, the price of each yard increasing in arithmetical progression; how much did the whole amount to? Ans. £5 2s. 3. How many times does the hammer of a clock strike in 12 hours? Ans. 78 times. NOTE. It is supposed that every scholar knows the first term and common difference of the hammer of a clock's striking. CASE III.-The extremes and number of terms being given, to find the common difference. RULE.-Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference. EXAMPLES. 1. A merchant sold 200 yards of cloth; for the first yard he received 4 cents, for the last, 601 cents; the common difference is required. Ans. 3, common difference. 200 601 price of the last yard. 199) 597(3 commmon difference. 597 DEM.-It is plain, that the difference between the price of the first yard, and the price of the last yard, that is, the difference of the extremes, is equal to the additions which were made to the price of the first yard till it became equal to the last yard, and these additions were one less than the number of terms, therefore it is evident, that the sum of these additions, that is, the difference of the extremes, divided by the number of terms, less one, that is, the number of additions, gives the common difference. 2. The first term is 3, the last term 48, and the number of terms 10; what is the common difference? Ans. 5. 3. A gentleman has 7 sons, whose several ages differ alike; the youngest is 5 year sold, the eldest 47; what is the common difference of their ages? Ans. 7 years. QUESTIONS ON ARITHMETICAL PROGRESSION. What is Arithmetical Progression? A. It is any rank or series of numbers increasing or decreasing by a common difference. What are numbers called when they increase by a continual addition? A. An ascending series. What are numbers called when they decrease by a continual subtraction? A. A descending series. When the first term, the common difference, and the number of terms are given, how do you find the last term? A. By multiplying the number of terms, less one, by the common difference, and adding the first term to the product for the last term. When the first term, common difference, and number of terms are given, how do you find the sum of all the terms? A. Multiplying one half the sum of the extremes by the number of terms, gives the answer or sum of all the terms. How do you find the common difference, when the extremes and number of terms are given? A. Dividing the difference of the extremes by the number of terms, less one, gives the common difference. GEOMETRICAL PROGRESSION, Is any rank or series of numbers increasing by one common multiplier as 2, 4, 8, 16, 32, &c. or decreasing by a common divisor; as, 32, 16, 8, 4, 2, &c. First, the multiplier, 2, by which the series is increased, is called the ratio; secondly, the divisor, 2, by which the series is diminished, is called the ratio. When any number of terms is continued in Geometrical Progression, the product of the two extremes will be equal to the product of any two means equally distant from the extremes, or when the terms are odd, equal to the square of the middle term; thus, 2, 4, 8, 16, 32; 2x32-64, 4×16-64, 8×8=64. There are five terms given in Geometrical Progression, the same as in Arithmetical Progression, viz: 1. The first term. 2. The last term. 3. The number of terms. NOTE. The first and last terms are called the extremes. Any two terms equally distant from the extremes, are called the means. Any three of these being given, the other two may be found. The powers of the ratio are frequently denoted by figures placed o ver them, called indices: 1 2 3 4 5, &c. Indices or Arithmetical scrics. Geometrical series. Thus, (2, 4, 8, 16, 32, &c. CASE I.-The first term, ratio, and number of terms being given, to find the last term. RULE.-Raise the ratio to a power whose index shall be one less than the number of terms; multiply that power by the first term, and the product will be the last term or answer. Or, write down a few leading powers of the ratio with their indices over them.-Add the inost convenient indices, to make an index one less than the number of the term sought.-Multiply together the powers belonging to those indices, and their product multiplied by the first term, will be the answer or term sought. EXAMPLES. 1. A merchant bought 9 yards of broadcloth, and by agreement, was to give what the last yard would come to, reckoning 4 cents for the first yard, 12 cents for the second, 36 cents for the third, and so on, to the last; what did the cloth cost him paying only the price of the ninth yard? Ans. $262,44 cents. 3×3×3×3×3×3×3×3×4-26244. Ans. Or thus: 1st power, 2d p. 3d p. 4th p. Ratio 3, 9, 27, 81 81 81 648 DEM. The reason of the rule is evident from the manner in which a geometrical series is formed, because in the geometrical series, 2, 4, 8, 16, 32, &c. which has already been given, the ratio being 2, it is plain, that the second term is formed by multiplying the first term by the ratio, the third term, by multiplying the second term by the ratio, and so on; hence it is obvious that the ratio raised to a power one less than the number of terms, and that power multiplied by the first term, must give the last term, because the same multipliers are used, and whether the first terin be first or last used asa factor, can make no difference. 8th power. 6561 4 Ans. or 9th term. 26244 2. A sum of money was divided among 8 persons, the first received $5, the second $15, the third $45, and so on, in three fold ratio; the share of the eighth person is re'quired. Ans. $10935, the share of the last. 3. A farmer bargains for 16 sheep, to pay only the price of the last, reckoning one cent for the first, two cents for the second, four for the third, and so on, doubling the price to the last; how much must he pay for them? Ans. $327,68cts. CASE II-The first term, the last term, and the ratio being given, to find the sum of all the terms or series. RULE.-Divide the difference of the extremes, that is, the difference of the first and last terms, by the ratio, less 1, and the quotient, increased by the greater term, will give the answer or sum of all the terms. EXAMPLES. 3645 the last term. 5 the first term. 2)3640 1. If the first term be 5, the last term 3645, and the ratio 3; what is the sum of all the the terms? Ans. 5465. DEM. The reason of this rule is obvious from the nature of a geometrical series; thus, if the ratio of any geometrical series be 2, the difference of the greatest and least terms is equal to the sum of all the terms except the greatest; let 2, 4, 8, 16, 32, be a geometrical series, whose ratio is 2, and whose sum is 2-4-4-4-84-164-32-62, and the difference of the extremes 32-2-30, and it is plain that the difference of the extremes is equal to the sum of au e omg except the last, for 2+4+8+16= 1820 3645 Ans. 5465 sum of all the terms. 30; the last term added to the difference of the extremes makes 322=30+32=62: Here dividing the difference of the extremes by the ratio, less one, could make no difference, because it would be dividing by 1; hence when the ratio is 3, the difference of the extremes, is double the sum of all the terms except the greatest; and when the ratio is 4, the difference of the extremes, is triple the sum of all the terms except the last, and so on. 2. The daughter of a wealthy gentleman, marrying on a new year's day, he gave her a guinea, promising to tripple it on the first day of each month in the year for her portion ; what was the daughter's portion? Ans. 265720 guineas. 3. What debt can be paid in a year, by paying one cent the first month, ten cents the second, and so on, increasing each month in a tenfold proportion? Ans. $1111111111,11 cents. NOTE,-First find the last term as in Case I, then find the sum of the series. QUESTIONS ON GEOMETRICAL PROGRESSION. What is Geometrical Progression? A. It is either increasing series of numbers by one common multiplier, or diminishing a number by one common divisor. When the first term, the ratio, and number of terms |