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2. How much water must be mixed with 100 gallons of rum, worth 7s. 6d. per gallon), to reduce it to Es. Sd. per gallon?

Sus. 20 gallons. 3. A fariner weuki mix 20 bushels of rye, at 65 cents per bushel, with barley at 51 cts. and oats at 30 cts. per bushel ; how much barley and oats inust be mixed with the 20 bushels of rye, that the provender may be worth 41 cents


bushel? Ans. 20 bushels of barley, and 614 bushe's of oats. 4. With 95 gallons of rum at 8s. per gallong I mixcil other rum at Os. 8d. per gallon, and some water; then I found it stood me in os. 44. per gallon; I demand how much rum and how much water I took ? Ans. 95 gals, rum et s. 8.1, and so gals, water.


CASE ini.


When the whole composition is loniteul to a given quantity:


Place the difference between the mean rate, and the several prices alternately, as in CASE 1.; then, As the sum of the quantities, or differencc 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 grocon had four sorts of tea, at 1s. Os. Gs. and 103. per lbs. the worst would not sell, anıl the best were too lear; he therefore mixed 120 in. and so much of each sort, as to sell it at 4s. per lb.; how much of each sort diil, he tahie ? lb.


76: 60 at 17
21. "lu.

2 : 20
I Aš 1 %: 120 : :


1.: 10 6 10

3 : 30


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Sum, 19


2. Ilow 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 aloriled at 6

*60 cents per gallon : Ans. SS; 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 inust be taken ?

wins, 40 lb. at 8 cts. 40 at 16 cts, and 160 at 24 cis. 4. A goldsmith liau two sorts of silver bullion, one of 10 07., and the other of 5 oz. fine, and bas a mind to inis a pound of it so that it shall be 8 oz fine; how much of each sort inust he take ?

Ans. 4 of 5* oz. fine, and 7 of 10 oz. fine. 5. Branly at 35. od. and is. 9d. per vallon, is to le mixeil, so that a hhd. of 65 gallons may be sold for 121. 1.25.; hory many gallons must be taken of each ?

Ans. 14 guls. at 5s. 9d. and 49 gals. at 3s. 6d.


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ARITHMETICAL PROGRESSION. Any rank of numbers mare than two, increasing by common excess, or decreasing, by common difference, is said to be in Arithmetical Progression.. So {; &

8, 6, 4, 2, &c. is a descending arithmetical series : The numbers which form the series, are called the terms of the progression; the first and lastterns of winch are called tiie extremes.*

PROBLEMI. The first term, the last tern, nad the hunber of terius being given, to find the sum of all the terms.

series in progression includes fire parts, viz. the first term, last term, number of terms, cormon diference, anil sum of the series.

By having any three of these purts giren, the other two may be foun:l, which admits of a variety of Problems; but most of them are best understood by an alrebraic process, and are here virittech

RULE. Multiply the sum of the extremes by the number of terns, and half the product will be the answer.


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.

23-4-3=26 sum of the extremes.

Then 20x11:2=145 the Answer. 2. Ilow many strokes does the hanner of a clock strike, in twelve hours ?

Ans. 78 3. A merchant soli 100 yards of cloth, viz. the first yard for 1 ct. the second for 2 cts, the third for $ cts. &c. I deinand what the cloth came to at that inte

Ans. $50%. 4. A man bought 19 yards of linen in arithmetical progression, for the first yaril he gave Is. and for the last yd. 11. 17s. what did the whole come to ? Anis. f: 18 ls.

5. A draper sold 100 yds. of broadcloth, at 5 cts, for the first yarı, 10 cts. for the second, 15 for the thirdl, &c. increasing 5 cents for every yard : What did the whole amount to, and what did it average per yarı?

Ans. Amowit, 8252), and the average price is $2,52cls. 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 terris given, to find the common difference.

RULE. Divide the difference of the extremes by the number of terms loss 1, and the quotient will be the common difference.


1. The extremes are 3 and 29, and the number of terms 14, what is the common difference ?

23} Extremes.

Ans. 4 year's.

Number of terms less 1=15) 26(2. Ans.

2. A man had 9 sons, whose several ages differed alike, the youngest was 3 years old, and the oldest 55; what was the common difference of their ages ?

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, ij. creasing every day by an equal excèss, so that the last day's journey may be 43 miles : Required the daily increase, and the length of the whole journey?

Ans. The duily increase is 5, and the whole jeurney 207 miles.

4. A debt is to be discharged at 16 different payments (in arithmetical progression, the first payment is to be 141. the last iQol.: What is the common difference, and the sum of the whole debt ?

Ans. 5l. 148, 8d. common difference, and 91 2. the uhole debt.

PROBLEM III. Given the first term, last terny, and cominon difference, to

find the number of terms.

RULE. Hvide the difference of the extremes by the cominon difference, and the quotient increased by i is the number of terins.


1. If the extremes be 3 and 45, and the common difserence 2; what is the number of terms ? 1128. 22.

2. A inan going a journey, travelled the first day fire niiles, the last day 45 miles, and each day increased his journey by 4 miles; how many days did he travel, and how lar? Ans. 11 days, and the whole distance travelled 275 miles

GEOMETRICAL PROGRESSION, Is when any rank or series of numbers increased by one common multiplier, or decreased 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. Maltiply 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.


1. If the series le 2, 6, 18, 54, 102, 486, 1459, and the ratio 3, what is its sum total ?


2186 the insurer.

S2. The extremes of a geometrical series are ) and 65536, and the ratio 4 ; what is the sum of the series :


Ans. 87581. PROBLEM II. Given the first term, and the ratio), to find any other term


CASE I. When the first terin of the series and the ratio are eqnal.!

*As the last term in a long series of numhers is very tedious to be found by continual multiplicatiors, it will be necessary for the readier finding it out, to have a series of numbers in arithmetical proportion, called indices, whose comuion difference is 1.

When the first term of the seriesand the ratio are equal, the indices must begin with the unit, und in this case, ihe

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