which numbers also perfectly satisfy the problem understood algebraically. If, however, it is required that the terms of the progression be positive, the last value of x would be inapplicable to the problem, though satisfying the algebraic equation. Several of the following problems also have two solutions, if we admit negative values. Prob. 3. Find three numbers in geometrical progression such that their sum shall be 210, and the last shall exceed the first by 90. Ans. 30, 60, and 120. Prob. 4. Find three numbers in geometrical progression such that their sum shall be 42, and the sum of the first and last shall be 34. Ans. 2, 8, and 32. Prob. 5. Find three numbers in geometrical progression such that their continued product may be 64, and the sum of their cubes 584. Ans. 2, 4, and 8. Prob. 6. Find four numbers in geometrical progression such that the difference between the first and second may be 4, and the difference between the third and fourth 36. Ans. 2, 6, 18, and 54. Prob. 7. Find four numbers in geometrical progression such that the sum of the first and third may be a, and the sum of the second and fourth may be b. Prob. 8. Find four numbers in geometrical progression such that the fourth shall exceed the second by 24, and the sum of the extremes shall be to the sum of the means as 7 to 3. Ans. 1, 3, 9, and 27. Prob. 9. The sum of $700 was divided among four persons, whose shares were in geometrical progression, and the difference between the greatest and least was to the difference between the means as 37 to 12. What were their respective shares? Ans. 108, 144, 192, and 256. Prob. 10. Find six numbers in geometrical progression such that their sum shall be 1365, and the sum of the third and fourth shall be 80. Ans. 1, 4, 16, 64, 256, and 1024. CHAPTER XVII. CONTINUED FRACTIONS.-PERMUTATIONS AND COMBINATIONS. 336. A continued fraction is one whose numerator is unity, and its denominator an integer plus a fraction, whose numerator is likewise unity, and its denominator an integer plus a fraction, and so on. The general form of a continued fraction is 1 a+1 b+1 c+1 d+, etc. When the number of terms a, b, c, etc., is finite, the continued fraction is said to be terminating; such a continued fraction may be reduced to an ordinary fraction by performing the operations indicated. 337. To convert any given fraction into a continued fraction. m n Let be the given fraction; divide m by n; let A be the quotient, and p the remainder: thus, Divide n by p; let a be the quotient, and q the remainder: and so on, so that we have m =A+1 a+1 b+, etc. We see, then, that to convert a given fraction into a continued fraction, we proceed as if we were finding the greatest common divisor of the numerator and denominator; and we must, therefore, at last arrive at a point where the remainder is zero, and the operation terminates; hence every rational fraction can be converted into a terminating continued fraction. Ex. 1. Transform 114 into a continued fraction. Ex. 2. Transform 346 into a continued fraction. Ex. 4. Transform 21 into a continued fraction. Ex. 6. Transform 130 into a continued fraction. 338. To find the value of a terminating continued fraction. Ex. 1. Find the value of the continued fraction 1 2+1 3+1. Beginning with the last fraction, we have L Ex. 2. Find the value of the continued fraction Ex. 3. Find the value of the continued fraction 1 2+1 3+1 2+1 2+. Ex. 4. Find the value of the continued fraction 339. To find the value of an infinite continued fraction. Let the fraction be 1 a+1 b+1 c+, etc. An approximate value of this fraction is obtained by omitting all its terms beyond any assumed fraction, and obtaining the value of the resulting fraction, as in the previous article. b+1=(ab+1)c+a' (bc+1)d+b 4th approximate value, (ab+1)cd+ad+ab+1, etc. 340. The fractions formed by taking one, two, three, etc., of the quotients of the continued fraction are called converging fractions, or convergents. The convergents, taken in order, are alternately less and greater than the continued fraction. 1 The first convergent is too great, because the denominator α 341. When a fraction has been transformed into a continued fraction, its approximate value may be found by taking a few of the first terms of the continued fraction. Thus an approximate value of 14 is, which is the first term of its continued fraction. By taking two terms, we obtain 2, which is a nearer ap proximation; and three terms would give a still more accurate value. 1193 Ex. 1. Find approximate values of the fraction 532. 424. Ex. 2. Find approximate values of the fraction 114. |