2. How much water must be mixed with 100 gallons of rum, worth 7s. 6d. per gallon, to reduce it to 6s. 3d. per gallon? Ans. 20 gallons. 3. A fariner would mix 20 bushels of rye, at 65 cents per bushel, with barley at 51 cts. and oats at 30 cents per bushel ; l:ow much barley and oats must be mixed with the 20 bushels of rye, that the provender may be worth 41 cts. per bushel ? Ans. 20 bushels of barley, and 61 4 bushels of oats. 4. With 95 gallons of rum at 8s. per gallon, I mixed other rum at 6s. 8d. per gallon, and some water; then I found it stood me in 6s. 4d. per gallon ; I demand how much rum ind how much water I took ? Ans. 95 gals. rum at 6s. 8d. and 30 gals. water. CASE III. 1 When the whole composition is limited to a given quantity. RULE. Place the difference between the mean rate, and the several prices alternately, as in Case I. ; then, As the sum of the quantities, or difference thus determined, is to the given quantity, or whole composition : so is the difference of each rate, to the required quantity of each rate. EXAMPLES. 1. A grocer had four sorts of tea, at Is. 3s. 6s. and 10s. per lb. the worst would not sell, and the best were too dear; he therefore mixed 120 lb. and so much of each sort, as to sell it at 4s. per lb.; how much of each sort did he take? 1- 6 6 : 60 at 1 lb. 6 3:30 - 10 2. How much water at 0 per gallon, must be mixed with wine at 90 cents per gallon, so as to fill a vessel of 100 gallons, which may be afforded at 60 cents per gallon ? Ans. 33} gals. water, and 664 gals. wine. 3. A grocer having sugars at 8 cts. 16 cts. and 24 cts. per pound, would make a composition of 240 lb. worth 20 cts. per lb. without gain or loss; what quantity of each must be taken? Ans. 40 lb. at 8 cts. 40 lb. at 16 cts. and 160 lb. at 24 cts. 4. A goldsmith had two sorts of silver bullion one of 10 oz. and the other of 5 oz. fine, and has a mind to mix a pound of it so that it shall be 8 oz. fine; how much of each sort must he take? Ans. 4 of 5 oz. fine, and 7} of 10 oz. fine. 5. Brandy at 3s. 6d. and 5s. 9d. per gallon, is to be mixed, so that a hhd. of 63 gallons may be sold for 121. 12s. ; how many gallons must be taken of each? Ans. 14 gals. at 5s. 9d. and 49. gals, at 3s. 6d. ARITHMETICAL PROGRESSION. ANY rank of numbers more than two, increasing by common excess, or decreasing by common difference, is said to be in Arithmetical Progression. So \ 2,4,6,8, &c. is an ascending arithmetical series : The numbers which form the series, are called the terms of the progression ; the first and last terms of which aro called the extremes.* PROBLEM I. The first term, the last term, and the number of terms being given, to find the sum of all the terms. * A series in progression includes five parts, viz. the first term, last term, number of terms, common difference, and sum of the series. By having any three of these parts given, the other two may be found which admits of a variety of Problems; but most of them are best under stood by an algebraic process, and are here omitted. RULE.—Multiply the sum of the extremes by the number of terms, and half the product will be the answer. EXAMPLES. 1. The first term of an arithmetical series is 3, the last term 23, and the number of terms 11; required the sum of the series. 23+3=26 sum of the extremes. Then 26x11:2=143 the Answer. 2. How many strokes does the hammer of a clock strike in 12 hours. Ans. 78. 3. A merchant sold 100 yards of cloth, viz. the first yard for 1 ct. the second for 2 cts. the third for 3 cts. &c. I demand what the cloth came to at that rate ? Ans. $507 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave ls. and for the last yd. 1l. 175. what did the whole come to? Ans. £18 ls. 5. A draper sold 100 yards of broadcloth, at 5 cts. for the first yard, 10 cts. for the second, 15 for the third, &c. increasing 5 cents for every yard; what did the whole amount to, and what did it average per yard ? Ans. Amount $252,, and the average price is $2,52 cts. 5 mills per yard. 6. Suppose 144 oranges were laid 2 yards distant from each other, in a right line, and a basket placed two yards from the first orange, what length of ground will that boy travel over, who gathers them up singly, returning with them one by one to the basket? Ans. 23 miles, 5 furlongs, 180 yds. PROBLEM II. The first term, the last term, and the number of terms 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. The extremes are 3 and 29, and the number of termi 14, what is the common difference? 29 22 Extremes . Number of terms less 1=13)26(2 Ans. 2. A man had 9 sons, whose several ages differed alike. the youngest was three years old, and the oldest 35; whai was the common difference of their ages ? Ans. 4 years, 3. A man is to travel from New-London to a certain place in 9 days, and to go but 3 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 43 miles: Required the daily increase, and the length of the whole journey ? Ans. The daily increase is 5, and the whole journey 207 miles. 4. A debt is to be discharged at 16 different payments (in arithmetical progression, the first payment is to be 14. the last 1001.; What is the common difference, and the sum of the whole debt ? Ans. 5l. 14s. 8d. common difference, and 9121. the whole debt. PROBLEM III. Given the first term, last term, and common difference, to find the number of terms. RULE.-Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms. EXAMPLES. 1. If the extremes be 3 and 45, and the common difference 2; what is the number of terms ? Ans. 22. 2. A man going a journey, travelled the first day five miles, the last day 45 miles, and each day increased his journey by 4 miles ; how many days did he travel, and how far ? Ans. 11 days, and the whole distance travelled 275 miles, GEOMETRICAL PROGRESSION, IS when any rank or series of numbers increase by one common multiplier, or decrease by one common divisor; as, 1, 2, 4, 8, 16, &c. increase by the multiplier 2; and 27, 9, 3, 1, decrease by the divisor 3. PROBLEM I. The first term, the last term (or the extremes) and the ratio given, to find the sum of the series. RULE. Multiply the last term by the ratio, and from the product subtract the first term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms. EXAMPLES. 1. If the series be 2, 6, 18, 54, 162, 486, 1458, and the 31 ratio 3, what is its sum total ? 3 x 1458_2 2186 the Answer. 3-1 2. The extremes of a geometrical series are 1 and 65536, and the ratio 4; what is the sum of the series? Ans. 87381. PROBLEM II. Given the first term, and the ratio, to find any other term assigned.* When the first term of the series and the ratio are equal.t * As the last term in a long series of numbers is very tedious to be found ☆ by continual multiplications, it will be necessary for the readier finding it out, to have a series of numbers in arithmetical proportion, called indices, whose common difference is 1. † When the first term of the series and the ratio are equal, the indices must begin with the unit, and in this case, the product of any two termş is equal to that term, signified by the sum of their indices : |