(2) S=a+b+c+d+ +g+h+k+l. Multiply both members of (2) by r, and it then follows that (3) Srar + br+er+ dr+ Subtract equation (2) from (3), hence . +gr+hr + kr + lr. Substitute in formula (ii) arn-1 for 7 [1530, (i)], and obtain the formula sought This formula enables one to calculate S when the first term a, the ratio r, and the number of terms n, of a G. P. are given. 539. One can easily verify formula (iii) in 538 on observing that the quotient of TMTM - 1, or, what amounts to the same thing, of 1 r” by 1 1 by r that is, as the sum of the first n terms of the G. P. where the first term is a and the ratio r. 540. Discussion of the Preceding Formula.- Consider the formula (iii) S= a(rn — 1) If the ratior is greater than 1, " increases while n increases, and consequently S increases, which is evident a priori. But, moreover, n can always be taken so large that will be larger than any assigned quantity, and consequently S also becomes greater than any assigned quantity (Lemma I). If r is less than 1, the formula may be written as follows, which is equivalent to the difference between two fractions with positive numerators, namely, In this form it is seen that the sum of the first n terms of a G. P. is which varies as n varies. As n increases, the fraction arn 1. -r dimin ishes, and consequently S increases; since as n increases, less and a 1 r less is subtracted from the fixed quantity But, when n increases without limit, the variable fraction can be made arn 1-r less than any assigned quantity, since the denominator of the fraction 1 r is fixed and the numerator ar" can be made less than any assigned quantity (Lemma II). Since and since arn 1 -p can be made as small as is desired, it follows, therefore, that S becomes as near equal to a 1 as is desired. This result is expressed by saying that, as n increases without limit, the sum S approaches the limit a Thus, arn -r 1 - r / n = ∞0 541. Applications.-1. Find the sum of the first fifteen terms of the G. P., 2. Calculate the limit of the sum of the decreasing G. P., 3. Calculate the limit of the sum of the decreasing G. P., 1 2n-1 Since the progression is composed of as large a number of terms as is desired, 2"-1 can be made larger than any assigned quantity and therefore smaller than any assigned quantity. Hence it may be said that by taking n large enough, the sum of n terms of the series can be made to differ from 2 by as small a quantity as is desired. This is abbreviated into the following: The sum of an infinite number of terms of this series is 2. 542. THEOREM III.-In a decreasing G. P. continued to infinity, each term bears a constant ratio to the sum of all which follow it, the common ratio being supposed less than unity. Let the series be a + ar + ar2 + ar3 + . then the nth term is arn-1; the sum of all the terms which follow this is Therefore, the ratio of the nth term to the sum of all which follow it is. so that this ratio may have a given value k, we put 543. Recurring decimals are cases of what are called infinite geometrical progressions. Thus, for example, .5343434 . . 5 34 34 34 denotes + + + + 5 10 103 105 107 Here the terms after ratio is 1 102 34 constitute a G. P. of which the first term is 103' 10 and the common Hence it follows that the sum of an infinite number 34 of terms of this series is 103 5 34 the value of the decimal is + 10 990 The general rule for such ex amples will be discussed in the next section, 544. To Find the Value of a Recurring Decimal.—Let F denote the figures which do not recur, and suppose that they are n in number; let Q denote the figures which do recur, and suppose that they are m in number. Let S denote the value of the recurring decimal; then = (10n+m10") SFQ-F. But 10+10" 10"(10"-1); and 10"-1 expressed by figures in the usual way is composed of m nines. Hence follows the rule for finding the value of a recurring decimal: Subtract the integral number consisting of the non-recurring figures from the integral number consisting of non-recurring and recurring figures, and divide by a number consisting of as many nines as there are recurring figures followed by as many ciphers as there are non-recurring figures. express two relations between the five quantities, a, l, r, n, and S; the two relations enable one to calculate any two of these five quantities, when the other three are given. One is therefore led to ten different problems, according as the quantities comprised in one of the following ten groups are taken as unknowns: (a, 1); (a, r); (a, n); (a, S): (l, r); (n, S). 546. PROBLEM IV.--Insert n geometric means between two given numbers, a and b; that is, terms, such that a and b are the form a G. P. composed of n + 2 extreme terms of this progression. Since the term b is preceded by b = a (), (ఉ), a b. and hence the (n+1)th term is and finally the (n + 2)th term is a rn+1 a()" *The abbreviation G. M. is used for the phrase, geometrical mean. EXAMPLE.—Insert 5 G. M. between 2 and 1458. 547. Three quantities are said to be in harmonical progression when a ca— -b: b C. Any number of quantities are said to be in H. P.* when any three consecutive quantities are in H. P. 548. The reciprocals of quantities in harmonical progression are in arithmetical progression. 549. The property demonstrated in 548 is sometimes taken as the definition of an H. P. and the definition stated in 547 proved as property of an H. P. The second definition of harmonical progres sion may be stated thus: quantities are said to be in harmonical pro"gression when their reciprocals are in A. P. a The term harmonical is derived from a physical property of musical sounds. Suppose there is a set of strings of the same substance, and whose lengths are proportional to 1, 1, 1, 1, 4, and ; and let them be stretched tight with equal forces. If now any two strings are sounded together, the effect will be harmonious to the ear. 3! The property of an H. P. derived in 548 enables us to solve some problems related to harmonical progression, although there is no formula for the sum of any given number of quantities which are in H. P., a being the first term and c the last. 550. Insert n harmonical means between two given terms a and b. The meaning of this problem is that one is to find n + 2 terms in an H. P. of which a is the first term and b the last term. problem can be reduced to the following: Insert n arithmetical means between and a Hence the The abbreviation H. P. is used for harmonical progression and H. M. for harmonical mean. |