by the continued division of 3, or multiplication of ), are in Geometrical Progression. 483. In general, if a represents the first term of such a series, and r the common multiplier or ratio; then may the series itself be represented by a, ar, ara, ar3, art, &c., which will evidently be an increasing or 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 arn--1. n--1 484. Let be the last term of a geometric series, then larm-land p2--1, ; ..r= The num a ber of intermediate terms between the first and last is n·2; let n-2=m, then n-1=m+1, and r= ", which gives the following rule for finding any m-+1 α number of geometric means between two numbers; viz. Divide one number by the other, and take that root of the quotient which is denoted by m+1; the result will be the common ratio. Having the common ratio, the means are found by multiplication. 485. Let S be made to denote the sum of n terms of the series (Art. 483), including the first, then a+ar+are+ar3+ +ar-2+ar--1=S. Multiply the equation by r, and it becomes ar+ar2+ar3 + +arn--2+urn--1+arn=rS. Whence, subtracting the first of these equations from the second, observing that all the terms except a and ar destroy each other, we shall have dr"—a=rS—S=(r−1)S ; and .•. S= arn -a r -1 Or, by substituting I for the last term ar-1, . (1). as above rl- -a found, this expression will become S= ; -1 from which two equations, if any three of the quantities a, q, n, 1, S, be given, the rest may be found. Thus, from the second equation, a=rl—(r−1)S; r = S-a and l=(r−1)S+a S-1' In the formula (1), when r=1, we have S= a-a 0 1-10 Now, the value of the symbol 0 in this be Ο particular case, shall be equal to na; because the series a+ar+ar2 + ... ar2--2+arn--1, for r=1, comes a+a+a+a+ &c., and the sum of n terms of this series, is evidently equal to na; therefore S= arn -a Jn. 1 -=na. Or, since 1. -7 = r 1 r -1 (Art. 112)a.[--1+pn-2+pn--3 ax[1+r+r2+p3 (Art. 128) a X +r+1]= ... 1], which, in the case of r=1, becomes a.[1+1+1+ &c.], and the sum of n terms of the series 1+1+1+ &c. is evidently equal ton; therefore S=a. =axnan, as before. 486. When the common factor r, in the above series, is a whole number, the terms a, ar, ar2, arn-1, form an increasing progression; in which case n may be so taken, that the value of the sum (S) shall be greater than any assignable quantity. a, 487. But if r be a proper fraction, as a α a '' ' 1 the series will be a decreasing one, and the expres sion (Art. 485), by substituting for r, and changing the signs of the numerator and denominator, (Art. definitely great; and consequently, by prolonging the ar' series, S may be made to differ from by less r-1 than any assignable quantity. 488. Whence, supposing the series to be continued indefinitely, or without end, we shall have in that case, S= ar' ; which last expression is what some call the radix, and others the limit of the series; as being of such a value, that the sum of any number of its terms, however great, can never exceed it, and yet may be made to approach nearer to it than by any given difference. 489. If the ratio, or multiplier, r, be negative, in which case the series will be of the form have, for the sum of an indefinite number of terms Ex. 1. Find the sum of the series, 1, 3, 9, 27, &c. Ex. 2. Find three geometric means between 2 and and the means required are 4, 8, 16. Ex. 3. The first term of a geometrical progression is 1, the ratio 2, and the number of terms 10. What is the sum of the series? Ans 1023. Ex. 4. In a geometrical progression is given the greatest term =4458, the ratio =3, and the number of terms =7, to find the least term. Ans. 2. Ex. 4. It is required to find two geometrical mean proportionals between 3 and 24, and four geometrical means between 3 and 96. Ans. 6 and 12; and 6, 12, 24, and 48. Ex. 6. Find two geometric means between 4 and 256. Ans. 16, and 64. Ex. 7. Find three geometric means between and Ans., 1, 3. 9. Ex. 8. A Gentleman who had a daughter married on New-year's day, gave the husband towards her portion 4 dollars, promising to triple that sum the first day of every month, for nine months after the marriage; the sum paid on the first day of the ninth month was 26244 dollars. What was the Lady's fortune? Ans. 39364 dollars. Ex. 9. Find the value of 1+1+1+}+ &c. ad infinitum. Ex. 10. Find the value of 1+2+&+37 &c. ad infinitum. Ans. 4. Ans. 2. § III. HARMONICAL PROPORTION AND PROGRESSION. 490. Three quantities are said to be in harmonical proportion, when the first is to the third, as the difference between the first and second is to the difference between the second and third. Thus, a, b, c, are barmonically proportional, when a:c::a-b: b-c, or a : c :: b-a: c-b. And c, [since a(b—c)=c(a—b) or ab=(2a—b)c], is a third harmonical proportion to a and b, when c= ab Za-b 491. Four quantities are in harmonical proportion, when the first is to the fourth, as the difference between the first and second is to the difference between the third and fourth. Thus, a, b, c, d, are in harmonical proportion, when ad::a b: c-d, or ad::b-a: d-c. And d, [since a(c-d)=d(a—b) or ac=(2a-b)d]; is a fourth harmonical proportional to a, b, c, when d= ac 2a-b In each of which cases, it is obvious, that twice the first term must be greater than the second, or otherwise the proportionality will not subsist. 492. Any number of quantities, a, b, c, d, e, &c. are in harmonical progression, if a:c::a-bb-c; b:d::b-c:c-d; c:e::c-d: d-e, &c. 493. The reciprocal of quantities in harmonical progression, are in arithmetical progression. For, if a, b, c, d, e, &c. are in harmonical progression; then, from the preceding Article, we shall have bc+ab=2ac ; dc+bc=2bd; ed+cd=2ec, &c. Now, by dividing the first of these equalities by abc; the second by 1 1 2 1 bdc; the third by cde ; &c., we have, 1 2 1 1 + 2 = ; &c. d C C e d 1 1 1 b'c' d'e sum. += + α c b b Therefore, (Art. 464), -, &c. are in arithmetical progression. 494. An harmonical mean between any two quantities, is equal to twice their product divided by their For if a, x, b, are three quantities in harmonical proportion, then (Art. 490), a: b:: a-x: x-b ; ax-ab-ab-bx, and x= 2ab a+b' Ex. 1. Find a third harmonical proportional to 6 and 4. Let the required number, then 6 x 6-4 : 4-x; .. 24-6x=2x, and x=3. |
