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twice, the fourth 4 times, and the three last, 7 times. Now if we multiply the last term by the ratio, the product 16 will contain the ratio 8 times, once more than what is contained in the three last terms, but as the first term is to be included in the sum, we shall obtain the sum by multiplying the last term by the ratio, and subtracting the first term. Again, let it be required to find the sum in the series 2, 6, 18.

Operation.
2, 6, 1 8 The ratio in this exam-
3 ple is 3, and 54 contains

- the ratio 18 times; or in 4 other words, it contains 2 the ratio 16 times, and —– the first term 3 times. By 2) 5 2 taking the first term once -- from 54, the remainder Sum of terms, 2 6 contains the ratio 16 times, and the first term twice. As the two last terms contain the ratio but 8 times, were we to take twice the first term from 52, and di. vide the remainder, 48, by 2, one less than the ratio, the quotient would express the sum of the two last terms ; but if we divide 52 by 2, the quotient will be two more than when we divided 48 by 2, that is, the sum of all the terms. - - The last term of any increasing series multiplied by the ratio, contains the first term as many times as the last term contains an unit, and the sum of all the other terms once less, therefore multiplying the last term by the ratio, and dividing the difference between that product and the first term, will give the sum of

the series. If the series be decreasing the principle is reversed. *

RULE.

If the series be increasing, multiply the last term by the ratio, and divide the difference between that product and the first term by the difference between i and the ratio. But if the series be decreasing, mul

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tiply the first term by the ratio, and divide the difference between that product and the last term by the difference between 1 and the ratio.

.Note—when the ratio is less than one, the sum is found the same as though the series were increasing, although in reality the series decreases. .

7. Find the sum of the terms in the following series: 3, 9, 27, 81. Ans. 120. 8. The number of terms of a geometrical series is 8, first term 3, and the ratio 4 ; what is the sum of the series 2 Ans. 65535.

9. A certain gentleman gave his whole estate to six sons,—giving to the youngest $1000 ; to the second once and a half as much, and to the others, sums increasing by the same ratio; how much was the estate 3 - Ans. $20781,25.

Required the sums of the terms in the following eaamples. - - *

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When two or more couplets of numbers have equal quotients, or ratios, this equality is called proportion; or, when the antecedents of two or more couplets, are the same parts of their respective consequents, or are the same multiples of their consequents, the numbers are said to be in proportion; the terms themselves are called proportionals. Thus, in the proportion 2:4:: 8: 16, 2, the antecedent of the first couplet, is the same part of 4, its consequent, that 8, the antecedent of the second couplet, is of 16, its consequent; and the proportion, 81 : 27: : 9: 3, 81 is the same multiple of 27, that 9 is of 3. When the ratio between the consequent of the first couplet and the antecedent of the second, is the same number as the ratio between the antecedent and consequent of each couplet, the proportion is said to be continued ; but when this is not the case, the proportion is called discontinued.

The OREM,

When four numbers are in geometrical proportion, the product of the extremes is equal to the product of the two means. But when the number of terms is odd, the product of the extremes is equal to the square of the middle term. .

Illustration.—When four numbers are in geometrical proportion, if one extreme be 1; the proportional next to it will always be the same as the ratio ; consequently, multiplying the other mean by that which is equal to the ratio, gives the greater extreme ; and this product of the means is evidently equal to the product of the extremes; for multiplying the greater extreme by the less, when that is one, does not increase it.

In the proportion 3 : 9:: 27: 81, three times 81 is 243, the 5th power of 3 the ratio, and 27, the third power of the ratio, multiplied by 9, the second power, gives 243, the 5th power. Also, in the proportion 2 : 6:: 18 : 54, two times 54, the product of the extremes, is 108, which is the same as 6 times 18, the product of the means, since 54 contains 18 three times. The same is equally true when the proportion is discontinued. 2: 6 : : 7 : 21. In this proportion, 42, the product of the extremes, is made up of six 7's, which is equal to the product of 6 multiplied by 7, the two means. This will always be the case, for the fourth term always contains the third as many times as the second contains the first, or is the same multiple of it. These examples show, that the product of the extremes will always equal the product of the means. -

When three terms are in proportion, and one of the extremes is 1, the mean is the same as the ratio, and multiplying the mean by itself, gives the other extreme, which is also the square of the mean. In the proportion 1:4:: 4: 16, the product of the extremes and the product of the means, each equal the square of 4, the ratio.

PROBLEM I.

To find any number of geometrical means between two - - given eactremes.

RULE,

Divide the greater extreme by the less, and extract such a root of the quotient as is denoted by one more than the number of means required. Then, if the series he increasing, multiply the first term by the rootjust found, continually, but if the series be decreasing, divide the first term continually by the same root, the products or quotients will be the means required.

Illustration.—According to Illustration, page 276, the greater extreme of any number of proportionals, is made up of as many times of that power of the ratio denoted by one less than the number of terms, as the less extreme contains units, therefore dividing the greater extreme by the less, gives that power of the ratio for a quotient, and extracting such a root of this quotient as is denoted by one less than the number of terms, or, (which is the same thing,) one more than the number of means, gives the ratio. 1. Let it be required to find two geometrical means between 3 and 81. Here 81--3=27, and ov/27=3, the ratio; then the first term 3 × 3=9, the first mean, and 9x3=27, the second mean. • 2. Find 3 geometrical means between 5 and 80. - Means, 10, 20, 40.

3. If a merchant sell 7 yards of cloth, and receive

prices for the same in geometrical proportion, what does he receive for the 2d, 3d, 4th, 5th, and 6th yards, if he receive 1 cent for the first yard, and $7,29, for the seventh * - Ans. 3, 9, 27, 81, 243 cents.

PROBLEM II,

Having one eactreme and one or two means given to find the other extreme. * ,

Illustration.—As the product of two means is equal to the product of the two extremes, it may be considered as the product of the two extremes; and if the product of two numbers be divided by either of those numbers, the quotient will be the other.

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When two means are given, divide their product by the given extreme. If but one mean is given, divide the square of the mean by the given extreme.

Remark. The nature of the question will generally determine which term is the extreme. As both antecedents, that is, the first and third terms are either multiplied or divided by the ratio to produce their consequents, it will at once be seen whether the third term is greater or less than that which is demanded. If greater, the greater of the two other terms, will of course, be the first term or antecedent; but if less, the first term will be less than the second.

4. What is the 4th term of four proportionals, if the product of the two means be 676 and the first term 4 3 Ans. 169.

5. What number must be divided by 4. that the quotient shall be the same number as the quotient of

39 divided by 132 - Ans. 12. 6. 35 bears the same proportion to 37 as 9 to what number 2 - Ans. 94%.

7. Suppose two companies of men spend money in proportion to the number of individuals in each company, and one company to consist of 11, and the

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