14. Find the proper fraction that, if changed to a continued fraction, will have as quotients 1, 7, 5, 2. 15. Find a series of fractions approximating to 0.5236; approximating to 0.7854. 16. Find a series of fractions approximating to the continued fraction that has as quotients 7, 2, 1, 2, 6, 4; that has as quotients 1, 2, 3, 4, 5, 6. Scales of Notation. 604. The common mode of representing numbers is called the common scale of notation, and 10 is called its radix or base. 605. In the common or decimal scale every figure placed to the left of another represents ten times as much as if it were in the place of that other. 606. Instead of the radix number 10, any other integral number might be used as the base of a system of notation. 607. A given number can be changed from one scale to another scale. 608. Examples. 1. Express 6532 in the scale of 6. SOLUTION. The quotients and remainders of the successive divisions by 6 are as follows: 6 6532 6 1088 remainder 4. 6 181 remainder 2. 6 30 remainder 1. 5 remainder 0. Therefore, 6532 expressed in the scale of 6 is 50,124. 2. Change 50,124 from the scale of 6 to the scale of 8. Therefore, the number required is 14,604. Since 50,124 is in the scale of 6, each figure has six times the value it would have one place to the right. Hence, at the beginning we have to divide 6 × 5 + 0 by 8, and we get 3 for the quotient and 6 for the remainder. The next partial dividend is 6 × 6 + 1, or 37, and this divided by 8 gives 4 for the quotient and 5 for the remainder. The next partial dividend is 6 × 5 + 2, or 32, and this divided by 8 gives 4 for the quotient and 0 for the remainder; and so on. 3. Change 14,604 from the scale of 8 to the scale of 10. SOLUTION. 10 14604 10 1215 remainder 2. 10 101 remainder 3. 6 remainder 5. Therefore, the required number is 6532. 4. Add 56,432 and 15,646 (scale of 7). 56432 15646 105411 SOLUTION. The process differs from that in the decimal scale only in that when a sum greater than seven is reached, we divide by seven (not ten), set down the remainder, and add the quotient with the next column. 5. Subtract 34,561 from 61,235 (scale of 8). 61235 34561 SOLUTION. When the number of any order of units in the minuend is less than the number of the corresponding order in the subtrahend, we increase the number in the minuend by eight instead of ten as in the common scale. 6. Multiply 5732 by 428 (scale of 9). 24454 5732 428 51477 12564 25238 2712127 SOLUTION. We divide each partial product by nine, set down the remainder, and add the quotient to the next partial product. 7. Divide 2,712,127 by 5732 (scale of 9). 428 5732) 2712127 25238 17722 12564 51477 51477 The operations of multiplication and subtraction involved in this problem are precisely the same as in the decimal notation. difference is that the radix number is 9 instead of 10. EXERCISE 145. Change 4852 of the common scale to: The only 9. The common seale 54,231 of the scale of 6.10. The common scale 54,231 of the scale of 7, 11. The common seale 54,231 of the scale of 8. 12. The common scale 54,231 of the scale of 9 Perform the following arithmetical processes: 13. Add 67,814; 76,406; 88,718 (scale of 9). to the si CHAPTER XVIII. SERIES. 609. Series. A succession of numbers that proceed according to some fixed law is called a series. sive numbers are called the terms of the series. The succes 610. A series that ends at some particular term is called a finite series. A series that continues without end is called an infinite series. 611. The number of different kinds of series is unlimited; in this chapter we shall consider only Arithmetical Series, Geometrical Series, and Harmonical Series. Arithmetical Progression. 612. A series of numbers that increase or decrease by a common difference is called an Arithmetical Series or an Arithmetical Progression. Thus, the numbers 5, 8, 11, 14 form an arithmetical progression with a common difference 3; and the numbers 12, 10, 8, 6 form an arithmetical progression with a common difference 2. we find any term, as the 6th, by adding to the first term the product of the common difference by a number one less than the number of the term: 2 + (3 × 5), or 17. In the decreasing arithmetical progression we find any term, as the 7th, by subtracting from the first term the product of the common difference by a number one less than the number of the term: 50 — (4 × 6), or 26. Hence, 614. To Find Any Term of an Arithmetical Progression, Multiply the common difference by a number one less than the number of the required term. Add this product to the first term if the series is an increasing series; subtract this product from the first term if the series is a decreasing series. EXERCISE 146. 1. Find the seventh term of the series 3, 5, 7, etc. 2. Find the fifteenth term of the series 2, 7, 12, etc. 3. Find the sixth term of the series 2, 25, 33, etc. 4. Find the twentieth term of the series 2, 31, 41, etc. 5. Find the seventh term of the series 21, 19, 17, etc. 6. Find the twelfth term of the series 18, 171, 163, etc. 7. If the first term of a series is 5, and the common difference 21, find the thirteenth and eighteenth terms. 8. If the fourth term of a series is 18, and the common difference 3, find the seventh and eleventh terms. 9. If the fifth term of a decreasing series is 52, and the common difference 3, find the twelfth and eighteenth terms. 10. If the fourth term of a series is 14, and the twelfth term 38, what is the common difference? HINT. The difference between the fourth and twelfth terms is evidently eight times the common difference. Find the common difference in a series : 11. If the fourth term is 12 and the seventh term 27. 12. If the first term is 20 and the fourth term 40. 13. If the first term is 2 and the eleventh term 20. 14. If the third term is 7 and the eighth term 12. 15. If the first term is 1 and the fourth term 19. |