SOLtmONS TO THE EXERCISES.] 



MATHEMATICS. ALGEBRA. 



505 



Let x and ;/ denote the two digits ; then the number is 

 lOx + y ; and by the question 



x + y = 15, and ry + 31 = 10y + x. 

 From the first, x = 15 y, .-.(15 y)y + 31 = 

 15 y. , 



That is, 15y y- + 31 = 9y + 15, .*. y 2 Cy = 1C. 

 Completing the square, 

 y- 6y + 9=25,.-.y 3 = +5, .-. y = 3 + 5 = 8, 



H-.vico the digits are 7 and 8, and the number is 78. 



10. There is a certain number consisting of two digits. 

 The left-hand digit is equal to three times the right-hand 

 one ; and if 12 be subtracted from the number, the re- 

 iiiiiiiider will be equal to the square of the left-hand 

 digit. What is the number ? 



Let x and y represent the two digits ; then the number 

 is lOj; + y. And by the question, 



x = 3y, and lOx + y 12 = x 2 

 . . by substitution, 30y + y 12 = 9y 2 



Transposing, 9y- Sly = 12 

 .-.Rule II., 18y 31 = J ( 36 x J2 + 31 2 ) 



31 + 23 4 



= V529=+23.-. y =-- = 3, or r 



lo 



The fractional value must be rejected, . . z = 3y = 9 ; 



the number is 93. 

 Oifterwlie. As above, 



lOx + y 12 = x 2 , andie 3y = 0, 

 Multiplying the first by 3, 30z + 3y 36 = 3z 2 

 Adding the second, x 3y =0 



.-. Six 36 = 3z a 

 Transposing, 3-c 2 31* = 36, . . Rule II, Cz 31 



31 -4-23 

 = V ( 36 X 12 + 31 2 ) = + 23. . * = - 



henco the digits are 9 and 3, and the number is 93. 



ARITHMETICAL PROGRESSION. EXAMPLES FOB EXERCISE. 



PAGE 478. 

 1. Find the sum of sixteen terms of the series 1 + 2 



By the formula, S = Jii {2a + (TV l)d] , where o = 1, 



d = 1, and n = 1 6. 

 Consequently S = 8 {2 + 15} = 130, the sum required. 



2. Find the sum of fourteen terms of the series 4 + 3 

 + 2 + 1 + 1 2 <fcc. Here a = 4, d = 1, aud 

 n = 14 : hence 



S = in {2a + (-])*} 



= 7 {8 13} = 35, the sum. 



3. Sum the series 1 + 1J + 2.i + (fee. to twenty terms. 



Here a J, a = 1, and n = 20 : therefore 

 S= Jn{2a + (/t !)<*} 



= 10{1 + 19} = 200, the sum. 



4. Insert three arithmetical means between 2 and 0. 

 I5y the formula, I = a + (n l)d, where a = 2, n = 5, 



1 = 0. .-.0=2 + 4<i, .-.d = f -- J. 

 Hence, when the means are inserted, the series is 

 2, 1J, 1, J, 0. 



5. Insert five arithmetical means between J and j. 



I = o + (n 1XA 



. 



Hence, when the means are inserted, the series is 



i, i, J, . J *. i- 



0. Tlio first term is 5, and the fifteenth 47 : what is 

 the common difference ? 



'. 47 



-iH- 



7. How many terms of the series 12 + 111 + 11 + 

 10.J + <kc., must bo taken to make 55 1 

 vol. i. 



QJ __ * /O _l 



.'. 55= Jn{24 (n 



. . 220 = 49n n 2 , . '. n 2 49n = 220 

 . .RuleII.,p.474,2(t 49= /(*9 2 880)= ^1521= 



49 + 39 

 . . n = - = 5, or 44 



Hence, whether five terms, or forty-four terms of the 

 series be taken, the sum will be 55. 

 8. The first term of an arithmetical progression is 7 



3 

 and the common difference <j : required the ninth term. 



.-. J = 7_8X = 7 12= 5, the ninth term. 



9. The sum of eight terms of an arithmetical series is 

 2, and the common difference 1J ; required the first 

 term. 



S = Jn{2a + (n l)d} 

 .-. 2=4{2a 7XH},.-. l = 4a 21, 

 .. 4a = 22, . . a = 5J, the first term. 



10. The first term of an arithmetical series in 3.^ , 

 and the common difference \ : required the sum of 

 twenty-one terms. 



.-.S 



= 70 + 42 = 28, the sum. 



GEOMETRICAL PROGRESSION. EXAMPLES FOR EXERCISE. 

 PAGE 479. 



1. Required the sum of five terms of the series. 



{ = ar - ', S = ?i=p where a = 1, r = 4, n = 5, 



. . I = 256, . . S = 3 = 341, the sum. 

 2. Required the sum of eight terms of 



= ar" - l , S = ^-?, where a = 1, r = -> n = 8, 

 ***-!, 



85 



128' 



: 256 3 



3. Required the sum of ten terms of 1 + 2 + 4 + 8 + <fec. 



I = ai~ - x , S = 

 .-.Z = 512, .-.S 



rl a 



r 1' 

 1024 1 



', where o = 1, r = 2, n = 10, 



= 1023, the sum. 



4. What is the geometrical mean between 6 and 54 ? 



54 X 6 = 324, and V3 24 = + 18 > mean. 

 Hence when the mean is inserted, the progression is 

 either 6, 18, 54 or 6, 18, 54, 



the ratio in the former case being 3, and in the latter 

 3. 



5. Insert two geometrical means between 2 and 54. 

 When the two means are inserted there will be four 

 terms, therefore n = 4 ; also a = 2, and I = 54 ; and we 

 liave to find r. By the formula, I = ar" - 1 , . . 54 = 2r>, 



c. to infinity. 

 1 



Hence, supplying the means, the progression is 2, 6, 18, 



1, 1 ,1 

 6. Sum the senes j^ T ^ T ^ 



By the formula, 

 S = r-^-3-'- ~ 



r" 10 10 9'' 



7. The first term of a geometrical series is 3. the 



3 T 



