and the fraction in the denominator be omitted, the ratio becomes

a result which shows that 22 is the nearest ratio expressed by a fraction with a denominator under 100, and that is the nearest ratio expressed by a fraction with a denominator under 1000. 427. After the quotients have been found, the results may be written as follows:

1

3+

428. The successive approximate values of a continued fraction are found by beginning at the top and taking first one, then two, then three, and so on, of its parts. Thus:

429. In reducing the part of a continued fraction selected for an approximate value, begin with the last fraction.

Ex. Find the value of the continued fraction

EXERCISE LXXXV.

1. Convert 8, 297, 135 into continued fractions. 2. Find the approximate values of 29; 4; 181.

3. Find common fractions approximating to .236; .2361; 1.609.

4. Find common fractions approximating to .382; 1.732; .6253.

5. Find approximate values of 1; 9; J; III. 6. Find the proper fraction that, when reduced to a continued fraction, will have 2, 3, 5, 6, 7 as quotients. 7. Find a series of fractions approximating to the ratio of the pound troy (5760 grs.) to the pound avoirdupois (7000 grs.).

8. Find a series of fractions approximating to the ratio of the side of a square to its diagonal; that ratio being 1:1.414214 nearly.

9. Find a series of fractions approximating to the ratio of the ar to the square chain, from the equality 1 ar= .2471 of a square chain.

10. Find a series of fractions approximating to the ratio of the 48-pound shot to the weight of the French shot of 24kg

11. If the mean diameter of the Earth is reckoned at 7912 mi., and that of Mars 4189 mi., find a series of fractions approximating to the ratio of the mean diameters of these two planets.

12. Find a series of fractions approximating to the ratio of a cubic yard to a cubic meter from the equality 1 cu. yd. = .76453 of a cubic meter.

13. Find a series of fractions approximating to the ratio of the kilometer to the mile, from the equality

CHAPTER XXIV.

PROGRESSIONS.

ARITHMETICAL PROGRESSION.

430. A series of numbers that increase or decrease by a common difference is called an Arithmetical Progression.

Thus, the numbers 5, 8, 11, 14 are an arithmetical progression, since they increase by a common difference, 3; and the numbers 12, 10, 8, 6 are an arithmetical progression, since they decrease by a common difference, 2.

The several numbers of a series are called its terms.

431. In the series 2, 5, 8, 11, 14, 17, 20 it is obvious that the second term, 5, is 2+(3 × 1); the third term, 8, is 2+(3×2); the fourth term, 11, is 2+ (3 × 3); the fifth term, 14, is 2+ (3 × 4); and so on. Hence,

Any term may be found by multiplying the common difference by 1 less than the number of the term, and adding the product to the first term.

Thus, the twelfth term of the series 2, 5, 8..... will be 2 + (3 × 11) = 35.

In like manner, any term of a decreasing series will be found by subtracting this multiple of the difference from the first term.

Thus, the eleventh term of the series 50, 46, 42..... will be 50(10 × 4) = 10.

1. The seventh term of the series 3, 5, 7.....
2. The fifteenth term of the series 2, 7, 12.....
3. The sixth term of the series 2, 25, 34.....
4. The twentieth term of the series 2, 34, 4......
5. The seventh term of the series 21, 19, 17.....

6. The twelfth term of the series 18, 17, 16.....

7. When the first term of a series is 5, and the common difference 24, find the thirteenth and eighteenth terms.

Ex. When the fourth term of a series is 14, and the twelfth term 38, find the common difference.

The difference between the fourth and twelfth terms will evidently be eight times the common difference.

common difference will be 3814 – 3.

Find the common difference in a series:

8. Whose fourth term is 12 and seventh term 27.
9. Whose first term is 20 and fourth term 40.
10. Whose first term is 2 and eleventh term 20.
11. Whose third term is 7 and eighth term 12}.
12. Whose first term is 1 and fourth term 19.

Hence, the

432. The sum of seven terms of the series 3, 5, 7..... will be 3+ 5+ 7+9+11+13+15, or written in 15+13+11+ 9+ 7+ 5+ 3

reverse order,

Therefore,

twice the sum =18+18+18+18+18+18+18=18x7.

433. It will be seen that 18 is the sum of 3 and 15; that

is, the sum of the first and last terms; and that 7 is the number of terms. Hence,

The sum of an arithmetical series may be found by multiplying one-half the sum of the first and last terms by the number of terms.

Thus, the sum of eight terms of the series whose first term is 3, and last term 38, is

3. 8+7+7+ to sixteen terms.

4. 20+18+161+ to seven terms.

5. The first twenty natural numbers.

6. The natural numbers from 37 to 53 inclusive.

7. A series of thirty terms, of which the first is 21 and the last 59.

8. The series whose first two terms are 3 and 9 and last 75. 9. A series of twenty terms whose third and fifth terms are 10 and 15.

10. A stone, when dropped from a height, falls through 16.1 ft. in the first second, 48.3 in the next, 80.5 in the third, and so on, in arithmetical progression. How far will it fall in the seventh second? and how far in 7 sec.?

11. A, who travels 8 mi. the first day, 11 the second, 14 the third, and so on, overtakes in 17 dys. B, who started at the same time, and travelled uniformly. What is B's rate per day?

12. One hundred stones lie in a straight line, 1 yd. apart. A boy starts at the first stone, brings each of the others in separately, and piles them with the first stone. How far does he travel?