116 



THE POPULAR EDUCATOR. 



difference 5, and the number of terms 12, what is the last term ? 

 Ans. 3 = a-(7i-l)d=60-(12-l) X 5 = 5. 



(3.) If the first term of an ascending series be 9, and the 

 common difference 4, what will the 5th term be ? Ans. z = a 

 + (m-l) Xd = 9 + (5-l)x4= 25. 



There is one other inquiry to be made concerning a series in 

 arithmetical progression. It is often necessary to find the sum 

 of all the terms. This is called the summation of the series. 

 The most obvious mode of obtaining the amount of the terms is 

 to add them together. But the nature of progression will 

 furnish us with a more expeditious method. 



Let us take, for instance, the series 3, 5, 7, 9, 11. 

 and also the same inverted, 11, 9, 7, 5, 3. 



The sums of the terms will be, 14, 14, 14, 14, 14. 



Take also the series a a + cl, a + 2d, a + 3d, a -f- 4cl, 



and the same inverted, a + 4d, a + 3d, a + 2d, a + d, a 



The sums will be, 2a+4<Z, 2a+4d, 2a+4cZ, 2a+4cZ, 2a+4cZ. 



Hence it will be perceived that the sum of all the terms in 

 the double series is equal to the sum of the extremes repeated 

 as many times as there are terms. Thus, 



The sum of 14, 14, 14, 14, and 14 = 14 X 5. 

 And the sum of the terms in the other double series is 

 (2a + 4cl) X 5. 



But this is twice the sum of the terms in the single series. 

 If, then, we put 



a = the first term, n = the number of terms, 



z = the last, s = the sum of the terms, 



we shall have this equation, s = a "*" z X n. Hence 



2 



3. To find the sum of all the terms in an arithmetical pro- 

 gression. 



Multiply half the sum of the extremes into the number of 

 terms, and the product will be the sum of tlie given series. 



EXAMPLE. What is the sum of the natural series of numbers 

 1, 2, 3, 4, 5, etc., up to 1000 ? 



1 + 100 



Ans. s = 



2 



x 1000 = 500500. 



The two formulae, z = a + (n - l)cl, and s = 5 X n, con- 







tain five different quantities; viz., a, the first term; d, the 

 common difference ; n, the number of terms ; z, the last term ; 

 and s, the sum of all the terms. 



From these two formulae others may be deduced, by which, if 

 any three of the five quantities are given, the remaining two 

 may easily be found. The most useful of these formulae are the 

 following : 



By the first formula, 



1. The last term, z = a + (n- 1) d ; in which a, n, and cl are 

 given. 



Transposing (n - l)cl, 



2. The first term, a = z -f ( l)d ; z, n, and d being given. 



Transposing a in the first, and dividing by n 1, 



3. The common difference, d = 2 " 



Transposing and dividing, 



n 1 



; a, z, and n being given. 



4. TJie number of terms, n = -- + 1 ; a, z, and d being given. 



a 

 By the second formula, 



5. The sum of the terms, s = < ~^ X n a, z, and n being given. 



|| 



Or, by substituting for z its value, 



2a + (n 1)<Z . , . , 



s = X n ; m which a, n, and d are given. 



SB 



Reducing the preceding equation, 



6. The first term, a = 2 -<***+ 'to. g> d> and n being giyen< 



aft 



7. The common difference, d = -^ ; s, a, and n being 



given. n - n 



8. The number of terms, n = J(2a-d)*+8d,-2a + d . a> d> 



and s being given. %d 



A variety of other formulas may be deduced from the equa- 

 tions already given, the investigation of which will afford the 

 student a pleasing and profitable exercise. 



By the third formula, for example, may be found any number 

 of arithmetical means between two given numbers. For the 

 whole number of terms consists of the two extremes and all the 

 intermediate terms. If, then, m = number of means, m + 2 

 = n, the whole number of terms. Substituting m + 2 for n in 

 the third equation, we have 



The common difference, cl = z JH- a , in which a, z, and m are 



given. 

 EXAMPLE. Find 6 arithmetical means between 1 and 43. 



Here n = 8; a = 1 ; z = 43 ; d= Z a = 43 ~ l - 6, com- 



n1 8 1 



mon difference; /. the series is 1, 7, 13, 19, 25, 31, 37, and 43. 



It is obvious, from the mode in which we obtained an ex- 

 pression for the sum of an arithmetical series, that the sum of 

 the extremes is equal to the sum of any other two terms equally 

 distant from the extremes. Thus, in the series 3, 5, 7, 9, 11, the 

 sum of the first and last terms, of the first but one and last but 

 one, etc., is the same in each case, viz., 14. The same is true 

 of every series. 



EXEECISE 68. 



1. If the first term of an increasing arithmetical series is 3, the 

 common difference 2, and the number of terms 20, what is the sum of 

 the series ? 



2. If 100 stones are placed in a straight line, at the distance of a 

 yard from each other, how far must a person travel to bring them one 

 by one to a box placed at the distance of a yard from the first stone ? 



3. What is the sum of 150 terms of the series 



-. -, 1. -, 5 , 2, I, etc.? 

 3 3 33 3 



4. If the sum of an arithmetical series is 1455, the least term 5, and 

 the number of terms 30, what is the common difference ? 



5. If the sum of an arithmetical series is 567, the first term 7, and 

 the common difference 2, what is the number of terms ? 



6. What is the sum of 32 terms of the series 



1, 1$, 2, 2, 3, etc. ? 



7. A gentleman bought 47 books, and gaye 10 shillings for the first, 

 30 shillings for the second, 50 shillings for the third, etc. What did 

 he give for the whole ? 



8. A person put into a charity-bor a shilling the first day of tha 

 year, two shillings the second day, three shillings the third day, etc., 

 to the end of the year. What was the whole sum for 365 days ? 



9. How many strokes does a common clock strike in 24 hours ? 



10. The clocks of Venice go on to 24 o'clock ; how many strokes do 

 they strike in a day ? 



11. Required the sum of the odd numbers 1, 3, 5, 7, 9, etc., continued 

 to 100 terms ; and also to n terms. 



12. Required the 365th term of the series of even numbers 2, 4, 6, S, 

 10, 12, etc. ; and also the nth term. 



13. The first term of a series is 4, the common difference 3, and tho 

 number of terms 100. What is the last term, and als the nth term ? 



14. A man puts 1 out to interest at 6 per ceat. ; what will be the 

 amount in 40 years at simple interest ? 



15. The extremes of an arithmetical series are 2 and 29, and the 

 number of terms is 10. What is the common difference ? 



16. The extremes of an arithmetical series are 3 and S9, and the 

 common difference 2. What is the number of terms ? 



17. Find 5 means between 6 and 48. 



18. Find 6 means between 8 and 36. 



Problems of various kinds, in arithmetical progression, may 

 be solved by stating the conditions algebraically, and then 

 reducing the equations. Thus : 



EXAMPLE. Find four numbers in arithmetical progression, 

 whose sum shall be 56, and the sum of their squares 864. 



Let x = the second of the four numbers, 



And y = their common difference. 



The series will be x - y, x, x -f y and x -f 2y. 



By the conditions, (x y) + x + (x + y) + (x + 2>j] -s 56. 



And (x - y)* + x 2 + (x + y) 2 + (x + 2y) 2 = 864. 



That is, 4s + 2y = 56. 



And 4z 2 + 4xy + 6y 2 = 864. 



Reducing these equations, we have x = 12, and y = . 

 The numbers required, therefore, are 8, 12, 16 and 20. 



EXAMPLE. A certain number consists of three digits, which 

 are in arithmetical progression, and the number divided by the 



