LESSONS IN ALGEBRA. 



115 



to the amount of water whioh the man drinlu, but its omiuion 



. ' i moans destroys the sentence. 

 Tims, thi-ii, we hare the proposition reduced to this form : 



-088 of a similar kind, the proposition may bo 

 still i. 



Let r >d that you wish to have and contemplate the 



idea of wat-.-r bring drunk, in it* moat elementary form, tlu-n 

 not need the article the ; accordingly the proposition now 

 assumes this form, Man .'/-.,, /,,v fater. 



A third process of simplification brings the sentence to these 

 two words, Mun drinks, whioh set forth the simplest statement 

 you can make on the subject. Remove the word man, you have 

 no sense ; remove the word drinks, you have no statement. 

 [iiuntly the original proposition, when reduced to man 

 drinks, is in its simplest form. 



Such, then, is tho form to which all propositions or sentences 

 may be reduced. What does the form involve ? Here are two 

 Those two words you recognise as a noun and a verb, 

 the one denoting a being and the other an act. Being and 

 doing are the great facts witli which all science is concerned, 

 and the relation of being to doing, so far as the utterance of 

 that relation is concerned, is the affair of the grammarian. The 

 simplest proposition consists of a noun and a verb so related 

 that what the verb declares is declared of the noun which is 

 the subject of the proposition. 



Agreement. This, the simplest form of a proposition, may 

 undergo modifications. You may change tho subject : for in 

 stance, you may make the singular man into the plural men ; but 

 if you moke this change, you must also change the verb, substi- 

 tuting drink for drinks. Here you see an instance of gramma- 

 tical agreement. Man drinks, men drink ; these pairs of words 

 severally agree, but in man drink and men drinks tho pairs do 

 not agree. Hence yon learn that a singular noun requires a 

 singular verb, and a plural noun requires a plural verb. 



Agreement, then, is the grammatical correspondence of two or 

 more words one with another ; this kind of agreement, however, 

 is in English of less prevalence than another, which may bo 

 called logical. Grammatical agreement is an agreement in 

 form ; thus, in the above example, the question is whether or 

 aot, and in which case the s should bo added to the verb drink. 

 But when I say Hie sick man drinks, and declare that there is 

 an agreement between side and man, do I mean an agreement 

 of form f No, for sick remains unchanged whatever noun you 

 append to it; thus we say sick man, sick men. sick women, sick 

 toy, sick girls. The agreement, then, is not in form. Yet the 

 two words "sick man" do agree; in what? In sonso ; there 

 is a logical agreement. 



Observe, also, that the grammatical includes the logical, but 

 tho logical does not include tho grammatical. The last state- 

 ment has just been illustrated. The former may be shown to 

 be true, thus : when I say the man drinte, I make a statement 

 in whioh tho word man agrees in thought with the word drinks; 

 that is, the two go together ; the two are combined so as to 

 make a proposition ; the two agree in sense. 



Instances of Agreement. These are afforded in the following 

 table . 



THE SICK ) ,. 



(gram, agr.) t MAX \ * d J' and nouu ( lo 'J- a r -> 

 noun and verb 1 DRINKS \ 



COPIOUSLY. \ verb and adv. (log. agr.) 



Sentences may be either affirmative, negative, interrogative, 

 or interrogative negative : for example : 



Affirmative. I love my father. Interrog. Do I love my father' 



Negative. I do not love uiy father Int. Keg. Do Inot love mj father? 



LESSONS IN ALGEBRA. XXXVII. 



ARITHMETICAL PROPORTION AND PROGRESSION. 



IT four quantities are in arithmetical proportion, the sum of the 

 extremes is equal to the sum of the means. 



Thus, if a b . . h m, then a + m = b + h. 

 For by supposition, a - b = h - m. 



And transposing - 6 and - m, a + m = b + h. 



So, in the proportion, 12 -10: :11 -S 1 , we have 12+9 = 10+11. 



Again, if three quantities aw In arithmetical proportion, IKf 

 $um of the exlremei it equal to double the mean. 



It a -I : :b-c, then, a - b r= I - e. 



And transposing 6 and c, a + e = 2fc. 



Quantities which im-fease by a common difference, as 2, 4, 6, 

 8, 10, etc., or decrease by a common difference, as 15, 12, 9, 6, 3, 

 etc., are in continued arithmetical proportion. 



Such a series in also called an arithmetical progression ; and 

 sometimes progression by difference, or equidi/erent series. 



When the quantities increase, they form what is called mo 

 ascending series, as 3, 5, 7, 9, 11, etc. 



When they decrease, they form a descending series, M 11, 9, 7, 

 5, 3, etc. 



The natural numbers, 1, 2, 3, 4, 5, 6, etc., are in arithmetic*: 

 progression ascending. 



From the definition it is evident that, in an ascending series, 

 each succeeding term is found by adding the common difference 

 to the preceding term. 



If the first term is 3, and the common difference 2, 

 The series is 3, 5, 7, 9, 11, 13, etc. 

 If the first term is a, and the common difference d, 

 Then a + d is the second term, a + d + d = a+2d the third, 

 a + 2d + d = a + 3d the fourth, a + 3d + d = a+4dtke 

 fifth, etc. 



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



And the series is a, a + d, o + 2d, a + 3d, a + 4d, etc. 

 If the first term and the common difference are the same, the 

 series becomes more simple. Thus, if a is the first term, and 

 also the common difference, and n the number of terms, 

 Then a + a = 2a is the second term, 

 2a + a = 3a the third, etc. 



And the series is, a, 2a, 3a, 4a na. 



In a descending series, each succeeding term is found by sub- 

 trading the common difference from the preceding term. 



If a is the first term, and d the common difference, the 

 series is 1st. 2nd. 3rd. 4th. 5th. 



a, a - d, a - 2d, a - 3d, a - 4d, etc. 



In this manner we may obtain any term, by continued addition 

 or subtraction. But in a long series this process would become 

 tedious. There is a method much more expeditions. By attend- 

 ing to the series, 



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



a, a + d, a + 2d, a + 3d, a + 4d, etc., 



it will be seen that the number of times d is added to a is one 

 less than tho number of the term. Thus, 



The second term is a + d, i.e., a added to once d ; 

 The third is a + 2d, ,, a ,, to twice d ; 

 The/ouv<7i is a + 3d, a to thrice d, etc. 

 So if the series be continued, 



The 50th term will be a + 49d. 

 The 100th term a + 99d. 



If the series bo descending, the 100th torm will be a 99d. 



In the last term, the number of times d is added to a is on 

 less than the number of all the terms. 



If, then, d the common difference, a = tho first term, = 

 the last, n = the number of terms, we shall have in all cases, 

 z=a+_(n-l)Xd; that is, 



1. To find the last term of an ascending series. 



Add to ihefrst term the product of the common difference into 

 the number of tei-ms minus one, and the sum will be the last 

 tei'm. 



2. To find tho last term of a descending series. 



F.'om the first term subtract the product of the common differ- 

 ence into tho number of terms minus one, and the remainder witt 

 be tlie last term. 



N.B. Any other term may be found in the same way. For 

 the series may be made to stop at any term, and that may be 

 considered, for the time, as the last. 



Tims, the mth term = a 4- (m - 1) X d. 



EXAMPLES. 



(1.) If the first term of an ascending series is 7, the common 

 difference 3, and tho number of terms 9, what is tho last term ? 

 Ans. t = a + (n - 1) u = 7 + (9 - 1) X 3 = 31. 



(2.) If the first term of a descending series is 60, the common 



