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GEOMETRICAL PROGRESSION Is when any rank or series of numbers are increased by one common multiplier, or decreased by one common divisor, as 1, 2, 4, 8, 16, 32, &c. increased by the multiplier %; and 27, 9, 3, 1, decreased by the divisor 3.

When any number of terms is continued in geometrical progression, the product of the two extremes' will be equal to any two means, equally distant from the extremes, as 2, 4, 8, 16, 32, 64, where 64x2=128, and 32x4=128, and 8x16=128. When the number of terms are odd, the middle term, multiplied into itself, will be equal to the two extremes, or any two means, equally distant from the mean, as 2, 4, 8, 16, 32; then 8 x8=64, and 2 x 32 = 64, and 4 X 16 = 64.

The five things in arithmetical progression are to be observed here also.

1. The first term.
2. The last term.
3. The number of terms.
4. The common difference, or ratio.

5. The sum of all the terms. As the last term, in a long series of numbers, is very tedious to come at by continual multiplication, therefore, for the readier finding it out, there is a series of numbers made use of, called indices, beginning with an unit, whose common difference is one. Whatever number of indices you make use of, set as many numbers in geometrical proportion, as is given in the question under them, As 1, 2, 3, 4, 5, 6 indices.

2, 4, 8, 16, 32, 64 numbers in geometrical proportion. But if the first term in geometrical proportion be different from the ratio, the indices must begin with a cipher. As 0, 1, 2, 3, 4, 5, 6 indices.

1, 2, 4, 8, 16, 32, 64 numbers in geometrical proportion. When the indices begin with a cipher, the sum of the indices made choice of, must be always one less than the number of terms given in the question; for 1 in the indices is over the second term, and 2 over the third, &c. Add any two of the indices together, and that sum will agree with the product of their respective terms.

As in the first table of indices % + 5
Geometrical proportion

4 X 32 = 128
And in the second table 2 + 4 =
Geometrical proportion * X 16 = 64

M

6

In any geometrical proportion, proceeding from unity, the

ratio being known, to find any remote term, without producing all the multiplications attending the operation.

Rule. Find what figures of the indices, added together, will give the exponent of the term wanted; then multiply the numbers standing under such exponent into each other, and it will give the term required.

NOTE. When the exponent 1 stands over the second term, the number of exponents must be one less than the namber of terms.

EXAMPLES.

1. A man buys 14 pears, and is to pay 21 mills for the first, 5 for the second, 10 for the third, &c. in a double geometrical proportion to the last, and is to get the whole, if he pays what the last one came to. How much did the pears cost him?

0, 1, 2, 3, 4, 5 indices or exponents.
1, 2, 4, 8, 16, 32 quarter cents, number of ternos.
32 X 32 = 1024 the 10th power of the indices.

8 the 3d power.

Quarter cents 4)8192 the 13th power, being one less than 14.

Ans. 2048 price of the last pear, 20 dolls. 48 cts.

3 indices.

In any geometrical progression, not proceeding from unity,

the ratio being given, to find any remote term, without producing all the intermediate terms.

RULE. Find the last term as above, only take care that every product be divided by the first term.

2. A sum of money is to be shared among eight men; the first is to have 80 dollars, and the second 240, and so on, in triple proportion geometrically. What will the last have?

0, 1, 2,

80, 240, 720, 2160 number of terms. 2160 X 2160

58320 X 240=13996800---80-$174960 Ans. 8 + 3+1=7 one less than the number of terms. 3. A man made his will in the following manner: To the children's schoolmaster 50 dollars ; his youngest son was to have as much more as the schoolmaster, and each son to exceed the former by as much more. There were nine sons. What was the oldest son's legacy?

Ans. 25600 dollars,

80

The first term, the last term, or the extremes, and the ratio,

given to find the sum of the series. RULE 1. Multiply the last term by the ratio, and from the product subtract the first term; then divide the remainder by the ratio less one, and the quotient will be the sun of all the terms.

4. If the series in a geometrical question be 2, 6, 18, 54, 162, 486, 1458, and the ratio 3, what is the total amount?

3X 1458-2

-E2186 answer. 31

ROLE 2, for all such questions. Find the last term, if it is not given, by question 2nd; then subtract the first from it, and divide the remainder by the ratio less one, to the quotient of which add the greater, and that will be the sam required.

Operation of No. 4, by Rule 2. 1458-2=1456;3—1=728+1458=2186 answer, 5. A young man agrees with a farmer to work 12 days, provided he would give him one-fourth of a cent for the first day, one cent for the second, and four for the third, &c. in. quadruple proportion geometrical. What did his wages amount to in 12 days? Ans. 13981 dollars 1+ cent.

6. A man married his daughter on new-year's-day, and gave

her husband 20 cents towards her portion, promising, if they would do well, to double it on the first day of every month, for one year. What was her portion ?

Ans, 819 dollars. NOTE. În working the above question, when putting down the terms under the exponents or indices, say, 1, 2, 4, &c. and the number or sum of. your series will be fifths of dollars, which divide by 5, and you will have the answer, as 20 cents equal one-fifth of a dollar.

7. A man bought a horse, and by agreement was to give one grain of corn for the first nail, two for the second, four. for the third, &c. There were four shoes, and eight nails in each shoe. I demand how many bushels of corn he paid for said horse, and how much the corn came to, at 25 cents per bushel, allowing 500 grains to make a pint, and 64 pints to a bushel ?

S 134217+ bushels of corn.

Amount 33554 dolls. 25 cts.

Ans. {

8. A crafty young man agreed with a farmer, (ignorant in numbers.) to serve him 12 years, and to have nothing for his labour, but the produce of a grain of wheat for the first year, and that product to be sowed for the second

year, and so on from year to year, until the end of said 12 years. I demand what the young man got for his wheat, supposing the increase to be only in a tenfold proportion, at 80 cents per bushel, 7680 grains of wheat to a pint, and 64 pints to a bushel? Ans. 180844 dolls. 80 cts. rejecting remainders.

9, A merchant sold 30 yards of very rich silk velvet, at 2 pins for the first yard, 6 pins for the second, 18 for the third, &c. in triple proportion geometrical. I demand how much the velvet produced, when the pins were afterwards sold, at 400 for a cent? Ans, 5147278302 dolls. 35 cts.

10. The extremes of a geometrical series are 1, and 65536, and the ratio 4; what is the sum of the series?

65536x4=262144-1=262143:4-1=87381 Ans. Note. Always observe, when the first term, (in geometrical proportion) of the series and the ratio are different, that is, when the first term is either greater or less than the ratio, the indices begin the exponents or indioes with a cipher, but if the first term and the ratio be alike, begin your indices with an unit.

11. A man bought 13 geese at market, and was to give apples in pay, and to get all the geese for the quantity of apples the last goose would come to, beginning with the first goose at two apples, and doubling up to the last. How many apples did the last goose come to, as that was what he paid for the whole ? 1, 2, 3, 4, 5 indices.

The first term and the ratio are alike.
2, 4, 8, 16, 32 terras.
5+5+3=13th term. 32X32X8=8192 apples, answer.

To find the whole amount of apples.
The last term 8192X2=16381-2=16382--2-1=16382 answer.

12. A man sold 20 oxen, the first for 3 cents, the second for 9 cents, the third for 27 cents, and so on, in triple proportion geometrical. What did the man get for his oxen?

Ans. 52301766 dollars.

ALLIGATION Is said, by some authors, to be of two kinds, Medial and , Alternate, and by some of three kinds, Medial, Alternate and Partial.

Alligation Medial is when the quantities and prices of several things are given to find the mean price of the mixture compounded of those things.

Rule. As the whole composition is to its total valae, so is any part of the composition to its mean price.

To prove your work, find the value of the whole composition at the mean rate, and if it agrees with the total value of the several quantities at their own rates, the work is right.

EXAMPLES.

1. A grocer would mix 20 pounds of sugar, at 8 cents a pound, and 15 pounds at 10 cents a pound, with 30 pounds at 12 cents a pound. What is one pound of this mixture worth?

Ans. 10 cents 3+ mills. 20 pounds at 8 cents is 160 cents.

15 30

10 12

150 360

:

65

670 As 65 : 670 :: 1 10 cents 3+ mills, answer. 2. A grocer mixed 19 gallons of rum, at 1 dol. 20 cts. per gallon, and 40 gallons of whiskey at 80 cts. per gallon, and 12 gallons of cider at 60 cents per gallon, together. I demand what a gallon of this composition is worth ?

Ans. 87 cents 3+ mills. 3. A farmer mingled 20 bushels of wheat, at a dollar per bushel, and 36 bushels of rye at 60 cents per bushel, with 40 bushels of corn at 40 cents per bushel. I want to know the price of one bushel of this mixture? Ans. 60 cts.

4. Suppose 5 lbs. of gold, of 22 carats fine, 2 lbs. of 21 carats fine, and a pound of alloy, be melted together, what is the quality or fineness of this composition ?

Ans. 19 carats fine. 5. A farmer hath several sorts of wheat, viz. one sort at 90 cents, another at 80 cents, and a third at 70 cents per bushel. He would mix an equal quantity of each together. I demand the price of a bushel of this mixture ?

Ans. 80 cents.

ALLIGATION ALTERNATE Is when the rates of several things are given to find sucla quantities of them, as are necessary to make a mixture, which may bear a rate compounded.

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