In what case would B never overtake A?

(26-1)2

Ans. When a>

86

For instance, in the preceding example, if B had started one day later, he could never have overtaken A.

Prob. 7. A traveler set out from a certain place and went 1 mile the first day, 3 the second, 5 the third, and so on. After he had been gone three days, a second traveler sets out, and goes 12 miles the first day, 13 the second, and so on. After how many days will they be together?

Ans. In 2 or 9 days. Let the student illustrate this example by a diagram like the preceding.

Prob. 8. A and B, 165 miles distant from each other, set out with a design to meet. A travels 1 mile the first day, 2 the second, 3 the third, and so on. B travels 20 miles the first day, 18 the second, 16 the third, and so on. In how many days will they meet? Ans. 10 or 33 days.

GEOMETRICAL PROGRESSION.

328. A geometrical progression is a series of quantities each of which is equal to the product of the preceding one by a constant factor.

The constant factor is called the ratio of the series.

329. When the first term is positive, and the ratio greater than unity, the series forms an increasing geometrical progression, as

in which the ratio is 2.

2, 4, 8, 16, 32, etc.,

When the ratio is less than unity, the series forms a decreas- . ing geometrical progression, as

in which the ratio is

81, 27, 9, 3, etc.,

330. In a geometrical progression having a finite number of terms, there are five quantities to be considered, viz., the first

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.

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

7=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.

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

Hence

rss — arn, α.

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

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

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

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

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

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 obtained by continued multiplication.

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.

5. a, n, s,|r, l, ar—rs=a-s; l(s—l)n—1— a(s—a)n—1.

6.1, n, s, a, r, a(s-a)n-1-l(s-l)n−1; (s—7)pn—spm—1 — — 7.

:

EXAMPLES.

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

We have

7=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.

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, 42, 38, etc.

21

512.

Ans. 19373 8. Find the sum of 15 terms of the series 8, 4, 2, 1, etc. Ans. 152047.

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, √ab, Vab3. 12. Find the value of 1+1+1+1+, etc., to infinity.

13. Find the value of 1+1+1+2+, etc., to infinity.

16 64

Ans. 3.

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. 1.

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 1

the ratio is, and the sum

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