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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 6s. 3d. per gallon ? Ans. 20 gallons.

3. A farmer would 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 must be mixed with the 20 bushels of rye, that the provender may be worth 41 cents per bushel?

Ans. 20 bushels of barley, and 61

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 and how much water I took?

Ans. 95 gals. rum at 6s. 8d. and 30 gals. water.

CASE III.

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 1s. 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. ; kow much of each sort did he take ?

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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. 334 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 less; what quantity of each must be taken ?

Ans. 40 lb. at 8 cts. 40 at 16 cts. and 160 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. 43 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 12. how many gallons must be taken of each ?

12s.;

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 <

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The numbers which form the series, are called the terms of the progression; the first and last terms of which are 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 understood 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 termn of an arithmetical series is 8, the last term 23, and the number of terms 11; required the sum of the series.

23+326 sum of the extremes.

Then 26x11÷2149 the Answer.

2. How many strokes does the hammer of a clock trike, in twelve 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. $50.

4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave 1s. and for the last yd. Il. 178. what did the whole come to? Ans. £18 1s.

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, 52cts. 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 greund 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 dif. ference.

EXAMPLES

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

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Extremes.

Number of terms less 1=13)26(2 Ans.

2. A man had 9 sons, whose several ages differed alike. he youngest was 3 years old, and the oldest 35; wha 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 paymenn (in arithmetical progression,) the first payment is to be 147. the last 100l.: 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 dif ference 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 Journey by 4 miles; how many days did he travel and how far?

his

s. 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.

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 scries be 2, 6, 18, 54, 162, 486, 1458, and the ratio 3, what is its sum total ?

3×1458-2

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2186 the Answer.

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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.*

CASE I.

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

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