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3. Multiply together the terms of the geometrical series belonging to those indices, and the product will be the term required.

EXAMPLES. 1. The first term of a geometrical series is 2, and the ratio 2; required the 13th term.

1, 2, 3, 4, 5, 6, 7, indices.
2, 4, 8, 16, 32, 64, 128, leading terms.

Then 6+7=index to the 13th term.

And 64x128=8192 the answer. 2. A young man agreed with a farmer to work for him 11 years, with no other reward than the produce of one grain of wheat for the first year, allowing the increase to be tenfold, and that produce to be sowed the second year, and so on from year to year, until the end of the time; what is the sum of the whole produce, allowing 7680 grains to make a pint, and what does it amount to, at one dollar and fifty cents per bushel ?

Ans. 2260561 +bush., and $339084,19cts. + NOTE.—In such questions, you first find the last term by one of the cases in Problem 2, and then the sum of the whole series by Problem 1.

3. A rich miser thought 20 guineas apiece too much for 12 fine horses, but readily agreed to give 4 cents for the first, 16 cents for the second, 64 cents for the third horse, and so on, in fourfold proportion, to the last ; what did they come to at that rate, and how much did they cost per head, one with another ?

The 12 horses came to $223696,20cts., and the average price was $18641,35cts. per head.

Case 2.- When the first term of the series, and the ratio, are different ; that is, when the first term is

either greater or less than the ratio.* RULE.-1. Write down a few of the leading terms of the series, as before, and begin their indices with a cipher ; thus: 0, 1, 2, 3, &c.

* When the first term of the series, and the ratio are different, the indices must begin with a cipher, and the sum of the indices made choice of, must be one less than the number of terms given in the question ; because 1 in the indices stands over the second term; and 2 in the indices over the third term, &c.; and, in this · 2. Add together the most convenient indices to make an index, less by 1, than the number expressing the place of the term sought.

3. Multiply the terms of the geometrical series together, belonging to those indices, and make the product a dividend.

4. Raise the first term to a power whose index is one less than the number of terms multiplied, and make the result a divisor.

5. Divide the said dividend by the said divisor, and the quotient is the term required.

NOTE. If the first term of any series be unity, or 1, the term required is found by multiplying the terms of the geometrical series together which belong to those indices, without needing any division...

EXAMPLES. 1. Required the 12th term of a geometrical series, whose first term is 3, and ratio 2.

0, 1, 2, 3, 4, 5, 6, indices.
3, 6, 12, 24, 48, 96, 192, leading terms.
Then, 6+5=index to the 12th term.
And 192x96=18432=dividend.

The number of terms multiplied is 2, and 2-1=1 is the power to which the term 3 is to be raised; but the first power of 3 is 3=divisor; therefore

18432;3=6144, the 12th term. 2. A goldsmith sold 1 tb. of gold, at 2cts. for the first ounce, 8cts. for the second, 32cts. for the third, &c., in quadruple proportion geometrical; what did the whole come to P

Ans. $111848,10cts. 3. A man bought a horse, and by agreement was to give a farthing for the first nail, two for the second, four for the third, &c. There were four shoes, and eight nails in each shoe ;-What did the horse come to at that rate ?

Ans. £4473924 58. 3d. 3qrs.

case, the product of any two terms, divided by the first term, is equal to that term beyond the first, signified by the sum of their indices. Thus:0, 1, 2, 3, 4, &c., indices.

wo: /1, 3, 9, 27, 81, &c., geometrical series. Here, 4+3=7, the index of the 8th term.

81 X27=2187, the 8th term, or 7th beyond the Ista 4. Suppose a certain body, put in motion, should move the length of one barleycorn the first second of time, one inch the second, three inches the third second of time, and so continue to increase its motion in triple proportion geometrical'; how many yards would the said body move in the space of half a minute ?

Ans. 953199685623yds. 1ft. lin. 1 bar. ; which is no less than five hundred and forty-one millions of miles.

ALLIGATION. ALLIGATION teaches to mix several simples of different qualities, so that the composition may be of a middle quality; and is commonly distinguished into two principal cases, called Alligation Medial and Alligation Alternate.

ALLIGATION MEDIAL. ALLIGATION MEDIAL is the method of finding the rate of the compound, from having the rates and quantities of the several simples given.

Rule.-Multiply each quantity by its rate; then divide the sum of the products by the sum of the quantities, or the whole composition, and the quotient will be the rate of the compound required. .

EXAMPLES. 1. Suppose 20 bushels of wheat at 10s. per bushel, 36 bushels of rye at 6s. per bushel, and 40 bushels of barley at 4s. per bushel, were mixed together; what would a bushel of this mixture be worth ?

20x10=200
36 x 6=216
40x 4-160

'96 576(6s. Answer.

576 2. A composition being made of 5 pounds of tea at 7s. per pound, 9 pounds at 8s. Ed. per pound, and 144 pounds at 6s. 103d. per pound; what is a pound of it worth?

1. Ans. 78. 4 d. + 3. A goldsmith mixes 8 pounds 54 ounces of gold of 14 carats fine, with 12 pounds 84 ounces of 18; what is the fineness of this mixture ? Ans. 16 carats..

gallon ?

4. If with 40 bushels of corn at 45 per bushel, there are mixed 10 bushels at 6s. per bushel, 30 bushels at 5s. per bushel, and 20 bushels at 3s. per bushel ; what will 10 bushels of that mixture be worth? Ans. $7 16cts.

5. A grocer would mix 12cwt. of sugar at 10 dollars per cwt. with 3cwt. at 8 dollars per cwt. and Scwt. at 71 dollars per cwt. ; what will a cwt. of this mixture be worth?

Ans. 88 95cts. Omills.to 6. If 16 gallons of brandy at I dollar 25 cents, and 4 gallons of water, be niixed with 40 gallons of wine at 3 dollars per gallon ; what will the mixture be worth per

Ans. $2,33 cts. ALLIGATION ALTERNATE.. ALLIGATION ÁLTERNATE is the method of finding what quantity of any number of simples whose rates are given, will compose a mixture of a given rate; so that it is the reverse of alligation medial, and may be proved by it.

Rulb.-Write the rates of the simples in a column under each other.

Connect, or link with a continued line, the rate of each simple, which is less than that of the compound, with one or any number of those, that are greater than the compound; and each greater rate with one or any number of the less.

Write the difference between the mixture rate, and that of each of the simple, opposite to the rates, with which they are respectively linked.

Then, if only one difference stand against any rate, it will be the quantity belonging to that rate; but if there be several, their sum will be the quantity,

EXAMPLES 1. A merchant would mix wines at 14s. 15s. 19s. and 22s. per gallon, so that the mixture may be worth 18s. per gallon; whạt quantity of each must be taken ?

Or thus: 4 at 14s.

( 14- 1..... at 14s. 18 15)

18 ) 1577) 1+4=5at155 3 at 195. Ans.

19 ) 3+4=7a19s 22- 4 at 22s.

(22 3 ..... at 22s.

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Ans. 18 3

19

2. How much corn at 2s. 6d. 3s. 8d. 4s. and 4s. 8d. per bushel, must be mixed together, that the compound may be worth 3s. 10d. per bushel ?

Ans. 12 at 28. 6d. 12 at 3s. 8d. 18 at 4s. & 18 at 4s. 8d.

3. A goldsmith has gold of 18 carats fine, 16, 19, 22 and 24 ; how much must he take of each to make it 21 carats fine? - Ans. 3oz. of 16, 1oz. of 18, 1oz. of 19,

5oz. of 22, and 5oz. of 24 carats fine. 4. It is required to mix brandy at 80 cents, wine at 70 cents, cider at 10cts. and water together, so that the mixture may be worth 50cts. per gallon.

Ans. 9gals. of brandy, 9 of wine, 5 of cider & 5 of water. CASE 2.- When the whole composition is limited to a cer

tain quantity. RULE.--Find an answer as before by linking; then say, as the sum of the quantities, or differences thus determined is to the given quantity, so is each ingredient found, to the required quantity of each."

1 EXAMPLES.

1, How many gallons of water at Octs. per gallon, must be mixed with wine worth 60cts. per gallon so as to fill a cask of 100 gallons, and that a gallon may be afforded at 50 cts.?.

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831 Ans. 16gallons of water, and 834 of wine. 2. How much gold of 15, of 17, of 18 and 22 carats fine, must be mixed together to form a composition of 40 ounces of 20 carats fine?

Ans. 5oz. of 15, 17 and 18, and 25oz. of 22. 3. Brandy at 3s. 6d. and at 5s. 9d. per gallon, is to be mixed, so that a hogshead of 63 gallons may be sold for £12 125.; how many gallons must be taken of each ?

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

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