

MATHEMATICS. ALGEBRA. 



[SOLUTIONS TO THE 



common ratio 6, and Uie but terra 375 : &ud tli 



urn. 



8rl a . a 1875 3 

 - j, . '. S - - - - _ 468, tho sum. 



8. Tho firt term of geometrical tenet u, and tho 



common tmtio ? : find the turn to infinity. 

 o 



By tho formula, 2 - -2, .: S - \ 



J- - 2^, the 



cum. 



1 2 

 9. Insert three geometrical means between -and -. 



When the means are inserted there will bo five terms, 



therefore n - 6 ; also a 



* 



find r. 



By the formula, I - r" -',/.?= ~r*, 



I, and* = ? ; and we have to 



Hence tho progression is 



1 1 , , 2 ,1,1 / , 2, 2 

 5, g V +3, +3, +3 V 3 g- 



10. Required the sum of the infinite series 



and then deduce 2 f or x = 2, and x = 2, 

 1 r ' x ' x 



Whenx=2, 2 = 2; when* = 2, 2=5. 



QUESTIONS IN WHICH PROPORTION AND PROGRESSION 

 ARE CONCERNED. PAGE 480. 



1. Divide 49 into two parts, such, that the greater 

 increased by C, may be to the less diminished by 11, as 

 9 to 2. 



Let x bo the greater, then 49 x is the less, and by 

 the question 



r + 6:38 x ::9 : 2. 

 Multiplying extremes and means, we have 



2z + 12 = 342 9x, . . llx = 330, .'. x = 30, 

 . . 49 x = 19 : henco tho two parts are 30 and 19. 



2. Two hundred stones are placed in a straight Hue, 

 at intervals of 2 feet, the first stone being 20 yards in 

 advance of a basket. Suppose a person, starting from 

 the basket, collects the stones, and returns them, one by 

 one to the banket. How much ground does he go over ) 



Upon returning the first stone, he will have gone 40 

 yards ; upon returning tho second, 40 yards and 4 feet ; 

 upon returning the third, 40 yards and 8 feet, and so 

 on ; hence wo have to find the sum of the arithmetical 

 progression, following, each term being feet. 



120 + 124 + 128 + 132 + <kc., to 200 terms. 

 By the formula, S = }n {2a + (71 l)d} 



.-. 8 = 100 {240 + 199x4} = 103, 600 feet, 

 and 103,000 feet = 34,533$ yards = 19J miles 213J yards. 



3. Tho sum of four numbers in geometrical progres- 



sion is equal to 1 added to the common ratio ; and . _ 



IB the first term : required the numbers. 

 Let r bo the common ratio, then since 





; Hence the progression i ^7' "*" 7' 



+ that is to 



ay, either of tho two progressions, following, will fulfil 

 tho condition*, namely, 



or, 



1 I II 

 17' 17' J7' IT' 

 1 4 16 _64 

 if If If 17' 

 4. From two towns, 165 miles apart, A and B set out 

 to meet each other. A travels 1 mile the first day. 2 

 the second, 3 the third, and so on ; B travels 20 miles 

 the first day, 18 the second, 16 the third, and so on : in 

 how many days will they meet ? 



Suppose they meet in n days, then 

 1 + 2 + 3 + <fco., to n terms = $n {2 + (n 1)} 



= A's distance 



and20+18+16+<fcc., tonterms= Jn {40 2(n 1)} 

 e '5 distance 



.-. the sum of the distances =Jn {42 n + 1} = 165 

 . . 42n n j + tv = 330, . . n 1 43n = 330, 

 . . Rule II., page 474, 2n 43 -= N / (43* 1320) 

 = V 529 = + 23 



Hence they meet in 10 days. If it were a condition that 

 B is to continue travelling after he meets and passes A, 

 diminishing his distance by two miles daily, he will on 

 the eleventh day travel miles, and therefore on the 

 twelfth day 2 miles ; that is, ho will return back 2 

 miles, on the thirteenth day 4 miles, and so on. The 

 second value of n above, shows that in 33 days B will 

 in this way again come up with A, by overtaking him. 

 In these 33 days A will have proceeded directly onwards 

 to a distance of 561 miles, while B will have been follow- 

 ing him for a distance of 396 miles, for B's whole distance 

 is expressed by 396 ; and 561 396 = 165 : so that 

 the algebraic sum of tho two distances still makes 1 ij."> 

 miles ; and the only condition implied in the algebraical 

 statement of tho question is, that the sum of the two 

 distances shall be 165 miles. 



5. The sum of the first and second of four numbers in 

 jeoinetrical progression is 15, and the sum of the third 

 and fourth 60 ; what are the numbers ? 



Let the geometrical progression be 



then, by the question, the conditions aro 

 | j + ? = 15 . . . (1), 



and x + xy-60 . . . (2), 

 ind since 4 times the first is equal to tho second, we 

 lave, after dividing each by x, 



J + 4 .1+ * _1 = , _ 4 



r, which is the same thing, -^ 1 = y(-, 1) 



/4 '\ 



hat is, transposing, (t/ + 1) (^ lj= ... (3). 



fow a product is 0, whiclievcr of tho factors ia ; hence 

 we must have either 



4 

 y + 1 - 0, or yi 1 = 0. 



The first of these conditions cannot be ad mi' 

 Mvaiiso it is plain, that for y = 1, the equation (2), 

 jecoming = 60, is impossible : hence tho condition 

 vhich is to satisfy (3) is 



1 



? 1 - 0, .-. 4 i/ = 0, .-. y s - 4, .-. v - + 2, 



-i?ri!i-*- 



lonce tho progression is either 



6, 10, 20, 60, or 15, 30, 60, 120. 

 6. The sum of three numbers in uoometrical progros- 

 ion is 35 ; and the mean term is to the difference of tho 

 xtremes as 2 to 3 ; what are the numbers 1 



