31. a= .2, s = 204.6, 7=102.4. How many terms of the series: ... 32. 4, 1, 12, will make a sum of 11? 33. 18, 6, 2, will make a sum of 364? GENERAL FORMULAS DERIVED FROM THE FUNDAMENTAL FORMULAS 417. From the fundamental formulas (I) and (II) in Arts 413 and 414 we may derive a formula for any desired element in terms of any other three elements. Illustrations: 1. Given a, l, and s; derive a formula for r. 2. Given a, n, and 7; derive a formula for s. NOTE. The general formulas for n involve logarithms, and are ordinarily reserved for advanced students. Exercise 129 1. Given r, n, and 7; 4. Given a, r, and derive a formula for a. derive a formula for a. THE INFINITE DECREASING GEOMETRIC SERIES 418. If the absolute value of r in a geometrical progression, a, ar, ar2, ....., arm-1, is less than 1, the successive terms become numerically less and less; hence, by taking a sufficiently great number of terms, that is, by making n sufficiently great, the nth term becomes as small as we may choose, although never equal to zero. Thus, if a = 1 and r = 1⁄2, the series, 1, 1, §, 1%, ..., may be continued until the nth term is so small as to have no assignable value. Hence, we may say that 0 is the limiting value of the nth term. It follows, therefore, that if I approaches 0, rl approaches 0. Hence, rl — a α - rl Hence, 8 = or approaches -r α as a limiting value. is the formula for the sum of the terms of an infinite decreasing geometric series. Illustration: Find the sum of the infinite series, 1+1+} +27+ 419. A recurring decimal is the sum of an infinite decreasing geometrical progression in which the ratio is 0.1 or a power of 0.1. .27+.0027+.000027 + is a geometrical pro gression in which a .27 and r = .01. Illustration: Find the value of .292929 ... .... In case the recurring portion of the decimal does not include the first decimal figures, the sum of the recurring portion is found as above and then added to the other. Thus: GEOMETRICAL MEANS 420. If we know a and 7, the first and last terms respectively in a geometrical progression, we may form a geometrical progression of m+2 terms by inserting m geometrical means between a and 7. Illustration: 1. Insert 4 geometrical means between 2 and We seek a geometrical progression of (m +2)=(4+2) = 6 terms. Hence, the required progression is 2, 3, 4, 7, 81, 743. 421. To insert a single geometrical mean, m, between the two numbers, a and 7, we require the value of m in The geometrical mean between two numbers equals the square root of their product. PROBLEMS INVOLVING GEOMETRICAL PROGRESSION 422. Illustrations: 1. The 3d term of a geometrical progression is 2, and the 7th term 162. Find the 5th term. Hence, the 5th term, art, = (?)(3)1 = (2) (3o) = 18. Result. 2. Find 3 numbers in geometrical progression, such that the sum of the 1st and 3d decreased by the 2d shall be 7, and the sum of the squares of the 3 shall be 91. 1. Find the first 3 terms of a geometrical progression whose 3d term is 9 and whose 6th term is 243. 2. Find the first 2 terms of a geometrical progression whose 5th term is and whose 10th term is 16. 3. Insert 4 geometrical means between 1 and 243. 4. Determine the nature, whether arithmetical or geometrical, of the series, 1, 1, 1, .... 5. Find the first 2 terms of a geometrical progression in which the 5th term is and the 12th term 16. 6. Which term of the geometrical progression, 3, 6, 12, is 3072 ? 7. Find to n terms the sum of the series, 1, 3, 9, 27, ... 8. Insert 4 geometrical means between 9. Find, to infinity, the sum of 2, 11⁄21⁄2, 1, .... and 32. 10. Find the value of the recurring decimal, 0.1515.... 11. Find the value of x such that x be in geometrical progression. · 1, x + 3, x + 11 may |