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term, the last term, the number of terms, the ratio, and the sum of the terms. When any three of these are given, the other two may be found. We will denote

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The first term and the last term are called the extremes, and all the other terms are called geometrical means.

331. In a geometrical progression, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms.

According to the definition, the second term is equal to the first multiplied by r, that is, it is equal to ar; the third term is equal to the second multiplied by r, that is, it is equal to ar2; the fourth term is equal to the third multiplied by r, that is, it is equal to ar3; and so on. Hence the nth term of the series will be equal to arm-1; hence we shall have

l=arn-1

332. To find the sum of any number of terms in geometrical progression, multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less one.

From the definition, we have

s=a+ar+ar2+.... +arn-3+arn-1

Multiplying this equation by r, we have

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Subtracting the first equation from the second, member from 'member, we have

rs-s=arn — ɑ.

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or, substituting the value of 7 already found, we have

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If we had subtracted the second equation from the first, we should have found

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which is the most convenient formula when r is less than unity, and the series is, therefore, a decreasing one.

333. To find the sum of a decreasing geometrical series when the number of terms is infinite, divide the first term by unity diminished by the ratio.

The sum of the terms of a decreasing series may be represented by the formula

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Now, in a decreasing series, each term is less than the preceding, and the greater the number of terms, the smaller will be the last term of the series. If the number of terms be infinite, the last term of the series will be less than any assignable number, and rl may be neglected in comparison with a. In this case the formula reduces to

α

S=

334. To find any number of geometrical means between two given

terms.

In order to solve this problem, it is necessary to know the ratio. If m represent the number of means, m+2 will be the whole number of terms. Hence, putting m+2 for n in the formula, Art. 331, we have

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That is, to find the ratio, divide the last term by the first term, and extract the root which is denoted by the number of means plus one. Having found the ratio, the required means may be ob tained by continued multiplication.

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contain five quantities, a, l, n, r, s, of which any three being given, the other two can be found. We may therefore have ten different cases, each requiring the determination of two quantities, thus giving rise to twenty different formulæ. The first four of the following cases are readily solved. The fifth and sixth cases involve the solution of equations of a higher degree than the second. When n is not large, the value of the unknown quantity can generally be found by a few trials. The four remaining cases, when n is the quantity sought, involve the solution of an exponential equation. See Art. 416. These different cases are all exhibited in the following table for convenient reference.

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(r−1)srn-
pn-1

n

71-1

1

n

1

n-1 2-1

-

l(s-7)n—1—a (s—a)n—1.

6.l, n, s, a, r, a(s—a)n—1—l(s—l)n−1; (s—1)μm — spn--1 — — l.

-1

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log. l-log. a

log. r

+1.

log. l-log. a

=

log. (s—a)—log. (s—7)+1.

log. [a+(r−1)s]-log. a log. r

log.l—log. [lr—(r—1)3] +1, log. r

EXAMPLES.

1. Find the 12th term of the series 1, 3, 9, 27, etc.

We have

l=arn-1=311=177147, Ans.

2. Given the first term 2, the ratio 3, and the number of terms 10; to find the last term.

Ans. 39366. 3. Find the sum of 14 terms of the series 1, 2, 4, 8, 16, etc.

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4. Find the sum of 12 terms of the series 1, 3, 9, 27, etc. Ans. 265,720.

5. Given the first term 1, the last term 512, and the sum of the terms 1023; to find the ratio.

6. Given the last term 2048, the number of terms 12, and the ratio 2; to find the first term.

7. Find the sum of 6 terms of the series 6, 41, 38, etc.

Ans. 1931 8. Find the sum of 15 terms of the series 8, 4, 2, 1, etc. Ans. 152043.

9. Find three geometrical means between 2 and 162. 10. Find two geometrical means between 4 and 256. 11. Find three geometrical means between a and b. Ans. Va3b, Vab, Vab3. 12. Find the value of 1+1+1++, etc., to infinity.

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13. Find the value of 1+++++, etc., to infinity.

16

Ans..

14. Find the value of 1+1+++, etc., to infinity. 15. Find the ratio of an infinite progression whose first term

is 1, and the sum of the series.

Ans. .

16. Find the first term of an infinite progression whose ratio is, and the sum .

Ans..

17. Find the first term of an infinite progression of which

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18. Find the value of the series 3+2++, etc., to infinity.

19. Find the value of the series

+1++, etc., to infinity. 20. A gentleman, being asked to dispose of his horse, said he would sell him on condition of receiving one cent for the first nail in his shoes, two cents for the second, and so on, doubling the price of every nail to 32, the number of nails in his four shoes. What would the horse cost at that rate? Ans. $42,949,672.95.

PROBLEMS.

Prob. 1. Find three numbers in geometrical progression such that their sum shall be 21, and the sum of their squares 189. Denote the first term by x and the ratio by y; then

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Transposing xy in Eq. (1), squaring, and reducing, we have x2+x2y2+x2y* =441–42xy.

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(3.)

Substituting this value of x in Eq. (1), and reducing, we have

Whence

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y=2 or, and x=3 or 12.

The terms are therefore 3, 6, and 12, or 12, 6, and 3.

Prob. 2. Find four numbers in geometrical progression such that the sum of the first and second shall be 15, and the sum of the third and fourth 60.

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Taking the first value of x and the corresponding value of y, we obtain the series 5, 10, 20, and 40.

Taking the second value of x and the corresponding value of y, we obtain the series -15, +30, −60, and +120;

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