Geometical Progression. 85When a series of numbers increase by a common multiplier, or decrease by a common divisor, those numbers are said to be in geometrical progression; such as 2, 4, 8, 16, &c. &c. The first and last terms are usually called the extremes, and the common multiplier or divisor the ratio. Note 1. If three numbers be in geometrical progression, the product of the two extremes will be equal to the square of the mean. Thus, if 3, 9, 27, be in geometrical progression, Then will 3X27=9X9. 2. If four numbers be in geometical progression, the product of the two extremes will be equal to the product of the means. Thus, if 2, 4, 8, 16, be in geometrical progression, Then will 2X16=4X8. 3. If a series of numbers, consisting of any number of terms, be in geometrical progression, the product of the two extremes will be equal to the product of any two means equidistant from the extremes; or to the square of the mean, if the terms be odd. Thus, if 1, 2, 4, 8, 16, 32, &c. be in geometrical progression, Then will 1X32=2×16=4X8. Or, if 1, 2, 4, 8, 16, &c. be in geometrical progression, 4. If, out of any series of numbers in geometrical progression, there be taken any series of equidistant terms, that series will likewise be in geometrical progression. Thus, if 2, 4, 8, 16, 32, 64, &c. be in geometrical progression, Then will 4, 16, 64, &c. be in geometrical progression. PROBLEM I. Given the number of terms the ratio, and either Then will 3+13=5+11=7+9.、 Or 1, 4, 7, 10, 13, &c. are in arithmetical progression, then will 1+13=4+10=7×2. 4. If, out of any series of numbers in arithmetical progression, there be taken any series of equidistant terms, that series will also be in arithmetical progression. Thus, if 2, 4, 6, 8, 10, 12, 14, &c. are in arithmetical progression, then will 4, 8, 12, &c. be in arith. prog. 5. In any series of numbers in arithmetical progression the common excess, or difierence, is as often repeated as there are terms in the progression, wanting one; viz. every term, except the first, is continually increased or diminished by the common excess or difference. 6. The rules for an ascending or descending series are the same; for a descending series becomes an ascending one by beginning at the least term. of the extreme terms, of a limited geometrical series, to find the other extreme. RULE. Write down a few terms of a geometrical series, beginning with, and formed by the given ratio; over which place the arithmetical series 1, 2, 3, 4, 5, &c. as indices; observe what figures of these indices, when added together, will give a number an unit less than that expressing the number of terms; and find the product of the terms in the geometrical series which stand under these indices. This product multiplied by the first term given in the question, or the first term divided by this product, according as the progression is increasing or decreasing, will give the term sought. PROB. II. Given one extreme, the ratio, and the number of terms, of a geometrical series, to find the sum of the terms. RULE. Find the other extreme by Problem 1. Then divide the difference between the extremes by the ratio less 1; the quotient increased by the greater extreme will give the sum of the terms. PROB. III. In any series of numbers in geometrical progression, decreasing ad infinitum, given the first term and the ratio, to find the sum of the series. RULE. Subtract the second term from the first; the square of the first term, divided by this difference, will give the sum of the series. g the greatest, the sum of the terms. r the ratio. n the number of terms. log logarithm of any letter. Then will the following theorems exhibit all the possible cases of geometrical progression, including those already given. 86. If a series of quantities increase or decrease by continual multiplication or division by the same quantity, then those quantities are said to be in geometrical progression. For the exponents of r are a series of numbers in arithmetical progression, beginning with zero, or nought, and increasing by unity in each succeeding term, as 0, 1, 2, 3; and therefore the exponent of the nth term is n-1; consequently, if the nth term be represented by 7, then will ar To find the sum of n terms of a geometrical progression, let 0 = first term, r the common ratio, n the number of terms, and s= the sum of (n) terms. Thus, the numbers 1, 2, 4, 8, 16, &c. which increase by continual multiplication by &c. which decrease by 1 1 1 2, and the numbers 1, ğ2 9' 27' continual division by 3, or multiplication by progression, as before. 87. In general, if a represents the first term, r the common multiple or ratio, then may the series itself be represented by a, ar, a2r, ar3, ar1, &c., which will evidently be an increasing or a decreasing_series, according as r is a whole number or a proper fraction. In the foregoing series, the index of r in any term is less by unity than the number which denotes the place of that term in the series. Hence, if the number of terms in the series be denoted by n, the last term will be ar. = S. 88. Lets be the sum of the series a, ar, ar2, ar3, &c.; then a + ar + ar2 + ar3 + &c. arn-2+ar-1 Multiply the equation by r, and it becomes ar + ar2 + ar3 + ar1 + &c... ... .... arm2 + arm + ar" =rs. Subtract the upper equation from the lower, or the series which is the value of s from the series which is the value of rs, and we have ar”—a—rs—s, or (r—1)s=ar”—a; and arn a r- -1° If r is a proper fraction, then r and its powers are less than 1. For the convenience of calculation, therefore, it is better in this case to transform the equation, into s= a-arn the numerator and denominator of the fraction 89. by multiplying If be the last term of a series of this kind, then 1 = ar11; .. rlar"; hence s= {ar T- -1 From this equation, therefore, if any three of the four quanti ties s, a, r, l, be given, the fourth may be found. rl-a Sa S= ; a=rl—(r—1)s; r= and l S value of n cannot be found from the equation by means of logarithms, thus: See page 193, Art. 92. Find the sum of the series 1, 3, 9, 27, &c. to 12 terms. Here a=1, r=3, n=12. be the last term of an arithmetic series, whose first term is a, common difference d, and number of terms n; then la+(n−1)d; .. (n-1)d—la, or d= a N 1° Now the num ber of intermediate terms between the first and the last is n -2; let n-2m, then n-1-m+1. Hence d a m+1' which gives the following rule for finding any number of arithmetic means between two numbers. Divide the difference of the two numbers by the given number of means increased by unity, and the quotient will be the common difference. Having the common difference, the means themselves will be known. See page 290. 91. The general expression for the sum of a geometric series whose common ratio (~) is a fraction, is s= a-arn Suppose now n to increase indefinitely; then 7" (r being a proper fraction) will decrease indefinitely; therefore ar" will decrease indefinitely with respect to a, or a will be the limit of a-ar", and press the value of the series when the number of its terms is supposed to be indefinitely increased, or, as it is commonly called, the sum of the series ad infinitum. See page 312. 92. If the sum of the series, the common ratio, and the first term be given, the number of terms may be found thus : rs—sta Since rs-sar"—a, or ar”—rs—sta, and 7”. a .. log. or nxlog. r=log. (rs-s+a)-log. a. Hence n= log. (rs-s+a)—log. a. log. r 1. The sum of a geometric series is 6560, its first term 2, and common ratio 3. What is the number of terms? 2. A person undertakes a journey of 364 miles, going one mile the first day, three the second, nine the third, and so on. will he arrive at his journey's end? Here s 364, a-1, r-3; ... N log. 728-log. 3 log. (rs-sa)-log. a 2.862728 6, days Ans. .477121 When Cubic Equations. = Let da3+cx2+bx N be any cubic equation, and let, as in the solution of quadratics x = r + y, or y + r = x, then by + br=bx cy2+2cry+cr2 = cx2 dy3 +3dry+3dr3y dr3=dx3 N dy3 + c'y2+ b'y +A=N Sum. c'z+ b's= b'y dz3 +3dsz2 + 3ds2z + ds2 = dy3 dz3 + c'z2 + c′′z + A′ = N' Sum. See page 144-5 Given bx+ax: = n Or dz3+e"z2 + b′′z = (N'—A') = N' Here it - N' Here it is evident that these equations are similar to the first,and thus we may successively find the values of r, s, t, &c. &c. the figures of the root. Now to find the denominators, put down the coefficients in a line thus, And if we suppose d=1, then the equation will be a +cx2+ bx M And then the values of r, s, t, &c. are as follows: Now, by substituting the values of c', b', c", b", &c. we have Hence it is evident how these formulas may be continued at pleasure, and all the quantities contained in the same are known; and since, in the second formula, 3r2+2cr+b form apart of the divisor, this may be used as a trial divisor for finding s, and so for the rest. An example will render these formulas very easy. |