LESSONS IN ' .'.',.. I ill A. 



247 



A ml tho series is r, r 2 , r 3 , r 4 , r*, r*, etc. 

 In :i / series, each succeeding term is found by 



ilii'iilint Uio preceding term by the ratio, or multiplying by the 



If the first term i-i ar*, and tho ratio r, 



.>* 1 



Tho second torrn is , or ar* X = ar 1 . 



r r 



And tho series is ar 8 , ar\ ar 4 , ar*, ar 1 , or, a, oto. 

 If tho first term is a, and tho ratio r, 



The series is a, , , , etc., or a, ar~, ar-*, eto. 

 r r* r* 



1st. 2nd. 3rd. 4th. 5th. 6th. 



By attending to tho sorios, a, ar, ar 2 , ar 3 , ar 4 , ar 8 , eto., it 

 will bo soon that, in each term, tho exponent of the power of 



i ID in one less than tho number of the term. 

 If then a = the first term, r = the ratio, 



z = tho last, n = tho number of terms, 



we have tho equation z = ar"' l , the last term ; that is, 



In irooraetrical progression, the last term is equil to the pro- 

 duct of the Jirst itito that power of the ratio whose index is one 

 less thnn tlw number of terms. 



When tho first term and tho ratio are the same, the equation 

 becomes z = rr* - l = r . 



Of the four quantities, a, z, r and n, any three being given, 

 tho other may bo found. 



1. By the last article, 



= ar*- l = the last term. 



2. Dividing by r* 1 - 1 , 



= a = the first term. 



3. Dividing the 1st by a, and extracting the root, 



(z \-i- 

 1 - 1 = r = the ratio. 



By the last equation may be found any number of geometrical 

 means between two given numbers. If m = the number of 

 means, ra + 2 = n, the whole number of terms. Substituting 

 m + 2 for n in the equation, we have, 



I + = r, the ratio. 



When the ratio is found, tho moans are obtained by continued 

 multiplication. 



The next thing to be attended to is the rule for finding the 

 sum of all the terms. 



If any term, in a geometrical series, be multiplied by the 

 ratio, the product will be the succeeding term. Of course, if 

 each of tho terms be multiplied by the ratio, a new series will bo 

 produced, in which all the terms except the last will be the 

 same, as all except the first in the other series. To make this 

 plain, let the new series be written under the other, in such a 

 manner that each torm shall be removed one step to the right 

 of that from which it is produced in tho lino above. 



Take, for instance, the series, 2, 4, 8, 16, 32. 



Multiplying each term by the ratio, 4, 8, 16, 32, 64. 



Here it will be seen at once that the last four terms in the 

 tipper lino aro the same as the first four in tho lower line. Tho 

 only terms which are not in both, aro the first of the one series, 

 and tho last of tho other. So that when we subtract the one 

 series from the other, all the terms except these two will dis- 

 appear, by balancing each other. 



If the given series is, a, ar, ar 1 , ar*, . . . ar" - * . 

 Then mult, by r, wo have ar, ar 2 , ar 3 , . . . ar"- 1 , ar*. 



Now lot s tho sum of the terms. 



Then, s = a + ar + ar 2 + ar 3 , . . . +ar"- 1 . 



And mult, by r, rs = ar -f- ar 8 -f- or*, . . . + ar" - T + ar*. 

 Subt. the first equation from the second, rs - s = ar a. 



And dividing by (r - 1), 



_ ar " a 



' ~f^T* 



In this equation, ar" is tho last term in the new series, and 

 is therefore the product of the ratio into tho last term in the 

 given series. 



" " ' that In, 



Therefore, i 



To find the Bum of a geometrical seriMt 



Multiply the latt term into the ratio, from the product lublraet 

 the firtt term, and divide the remainder by the ratio leu we. 



KXAMPLE. If in a, series of number* in geometrical progres- 

 sion, the firat term in G, tho hut term 1458, And the ratio 3, 

 what is tho sum of all the term* ? 



EXERCISE 71. 



1. Find tiro geometrical means between 4 and 256 



2. Find three geometrical mean* between j and 9, 



3. If the first term of a decreasing geometrical series is J. the ratio 

 J, and the number of term* 5, what is the sum of the series ? 



4. What is the sum of the series 1, 3, 9, 27, etc., to 7 terms f 



5. What is the sum of ten terms of the series 1, ], {. ,, etc. ? 



6. If the first term of a series is 2, the ratio 2, aud the number of 

 terms 13, what is the last term ? 



7. What is the 12th term of a series, the first term of which is o, 

 and the ratio 3 P Also find the sum of the series. 



8. A man bought a horse, agreeing to give one farthing- for the first 

 nail in his shoes, three for tho second, and so on. The shoes con- 

 tained 32 nails ; what was the cost of the horse ? 



Quantities in geometrical progression are proportional to their 

 differences. 



Let the series be a, ar, ar 2 , ar 3 , ar 4 , etc. 



By the nature of geometrical progression, 



a : ar : : ar : or* : : ar" . ar 3 : -. ar* : ar 4 , etc. 



In each couplet let the antecedent be subtracted from the 

 consequent. 



Then a : ar : : ar a : ar 1 ar : : ar 2 - ar : ar* ar 1 , etc. 



That is, the first term is to the second, as the difference 

 between the first and second to the difference between the 

 second and third ; and as the difference between the second and 

 third to the difference between the third and fourth, etc. 



If quantities aro in geometrical progression, their differences 

 are also in geometrical progression. 



Thus the numbers 3, 9, 27, 81, 243, etc. 



And their differences 6, 18, 54, 162, etc., 



are in geometrical progression. 



Problems in geometrical progression may be solved, as in 

 other parts of algebra, by means of equations. 



EXAMPLE. Find three numbers in geometrical progression, 

 such that their sum shall be 14, and the sum of their squares 84. 



Let the three numbers be x, y, and z. 

 By the conditions, x : y : : y : z, or xs = y s 



And x + y + z = 14. 



And x- + y* + z- = 84. 



From these three equations, *, y, and 2 =2, 4 and 8. Ans. 



EXERCISE 72. 



1. There are three numbers in geometrical progression whose pro- 

 duct is 61, and the sum of their cubes is 584. What are the numbers P 



2. There are three numbers in geometrical progression : the sum of 

 the first and last is 52, and the square of the mean is 100. What are 

 the aumbers ? 



3. Of four numbers in geometrical progression, the snm of the first 

 two is 15, aud the sum of the last two is 60. What are the numbers? 



4. A gentleman divided 210 amonf three servants, in such a manner 

 that their portions were in geometrical progression ; and the first bad 

 90 more than the last. How much had each ? 



5. There are three numbers in geometrical progression, the greatest 

 of which exceeds the least by 15 ; and the difference of the squares of 

 the greatest and the least is to the snm of the squares of all the three 

 numbers as 5 to 7. What are the numbers ? 



6. There are four numbers in geometrical progression, the second 

 of which is less than the fourth by 21; and the sum of the extremes 

 is to the sum of the means as 7 to 3. What are the numbers ? 



