GEOMETRICAL PROGRESSION. ANY series of numbers, the terms of which gradually increase or decrease by a constant multiplication or division, is said to be in Geometrical Progression. Thus, 4, 8, 16, 32, 64, &c. and 81, 27, 9, 3, 1, &c. are series in geometrical progression, the one increasing by a constant multiplication by 2, and the other decreasing by a constant division by 3. The number, by which the series is constantly increased or diminished, is called the ratio. PROBLEM I. Given the first term, the last term, and the ratio, to find the sum of the series. RULE.* Multiply the last term by the ratio, and from the product subtract the first term, and the remainder, divided by the ratio less 1, will give the sum of the series. EXAMPLES. * DEMONSTRATION. Take any series whatever, as 1, 3, 9, 27, 81, 243, &c. multiply this by the ratio, and it will produce the series 3, 9, 27, 81, 243, 729, &c. Now, let the sum of the proposed series be what it will, it is plain, that the sum of the second series will be as many times the former sum, as is expressed by the ratio; subtract the first series from the second, and it will give 729-1; which is evidently as many times the sum of the first series, as is expressed by the ratio less ; consequently 729-I 3-I sum of the proposed series, and is the rule; or 729 is the last term multiplied by the ratio, I is the first term, and 3-1 is the ratio less one; and the same will hold let the series be what it will. QE. D. NOTE EXAMPLES. 1. The first term of a series in geometrical progression is 1, the last term is 2187, and the ratio 3; what is the sum of the series ? NOTE I. Since, in any geometrical series or progression, when it consists of four terms, the product of the extremes is equal to the product of the means; and when it consists of three, the product of the extremes is equal to the square of the mean; it follows, that in any geometrical series, when it consists of an even number of terms, the product of the extremes is equal to the product of any two means, equally distant from the extremes; and, when the number of terms is odd, the product of the extremes is equal to the square of the mean or middle term, or to the product of any two terms equally distant from them. d by composition. : d by division. a a+b : c-d a+b :: c: c+d by conversion. a-b :: c+d: c-d mixedly. For in each of these proportions the product of the extremes is equal to that of the means. 2. The extremes of a geometrical progression are 1 and 65536, and the ratio 4; what is the sum of the series? Ans. 87381. 3. The extremes of a geometrical series are 1024 and 59049, and the ratio is 1; what is the sum of the series? Ans. 175099. PROBLEM II. Given the first term and the ratio, to find any other term assigned. RULE.* 1. Write down a few of the leading terms of the series, and place their indices over them, beginning with a cypher. 2. Add together the most convenient indices, to make an index less by 1 than the number expressing the place of the term sought. 3. Multiply the terms of the geometrical series together, belonging to those indices, and make the product a dividend. 4. Raise *DEMONSTRATION. In example 1, where the first term is equal to the ratio, the reason of the rule is evident; for as every term is some power of the ratio, and the indices point out the number of factors, it is plain from the nature of multiplication, that the product of any two terms will be another term corresponding with the index, which is the sum of the indices standing over those respective terms. And in the second example, where the series does not begin with the ratio, it appears, that every term after the two first contains some power of the ratio, multiplied into the first term, and therefore the rule, in this case, is equally evident. Y The |