478 



MATHEMATICS. ALGEBRA. 



[OKOMBTRICAL 1-ROOERSSIOJT. 



Ciren the firmt term (u) the com. diff. (J), and the 

 number of terms (n) to find the sum of i!. 



S-a + (a + i) f (a + SI) + (a + M) + ---- I, 

 or writing the term* in reverse order, 



/. adding, 23 - ( + + (" + + ( + + ( + 



+ ---- a + J. 



And M there are terms in each series for S, there 

 must bo n tqual term* in this last series ; that is to say, 



2S-n (a + S-l ( + ! but '- + < J X 



The foregoing expressions for I and S will enable you 

 to determine any two of the five quantities concerned 

 when the other throe are given ; as shown in the follow- 

 ing examples. 



1. The first term of on arithmetical progression is 5, 

 nml the ninth term 37 ; what is the common dill'oreuco 

 of tho terms f 



For tho particular numbers here given, the formula 

 I = a + (n l)d 



is 37=5 + 8d .-. d- 



~~. 



Therefore, the series is 5, 9, 13, 17. 21, 25, 29, 33, 37. 



_'. Required 1 + 3 + 5 + 7 + 9+ <tc., to twenty 

 terms. 



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



= 10 {2 + 19 x 2} =400. 



3. How many terms of tho series J + J + J- + <kc. 



(where <f = i) must be taken so that tho suui may be 

 nothing f 



1,1,1.. 1 11 



therefore, tho Borics i23G' ~C 3 2* 



4. Insert four arithmetical means betwojn 2 and 17. 

 Here 2 is tho first, and 17 the sixth term of an arith- 

 metical progression, and we have to find the common 

 difference. 



l=a + (n l)d 



17=2 + 5d .: a-=y=3 



.*. the means are, 5, 8, 11, 14 ; which evidently fill 

 up the gap between the first term and the sixth ; the 

 eric* being 2, 5, 8, 11, 14, 17. 



5. How many terms of tho series 9 + 7 + 5, <tc., 

 will give 21 1 



.-. 24 - Jn { 18 2(n 1) } = lOn n- 



.: tf lOn = 24 .-. 2n 10 = J1UO 9ti = +_2 

 .-. n = 4 or 6. 



Consequently, whether we take four terms of tho pro- 

 posed scries or fix, the gum in cither case will bo 24 ; 

 thus 



four lerou 9 + 7 + 5 + 3= 24. 



lit term* 9 + 7 + 5 + 3 + 1 1= 24. 



EXAMPLES FOB EXKBCISE. 



1. Find tho sum of sixteen terms of the series 



1 + 2 + 3 + 4 + &c. 



2. Find tho um of fourteen terms of tho scries 



4 + 3 + 2 + 1+0 1 2 Ac. 



3. Sum tho wries } + 1J + 2J + <tc., to twenty 

 t. . - 



4. Insert three arithmetical means between 2 and 0. 

 6. Insert five aritluuetical means between 1 and I. 



C. Tho first torm is 5, and tho fifteenth 47 ; what is 

 tho common diftMBMt 



ilnw many terms of tho Borios 12 + 11 \ + 11 

 + lOJ, io., must be taken to make 55 ? 



8. The first term of an aritlnmiir.il progression U 7, 

 the common differenco J : required the ninth torm. 



9. The sum of eight terms of an arithmetical Buries 

 is 2, and tho common dilTcronoo IV : required tho 

 first term. 



10. Tho first term of an arithmetical scries is 

 tho common dilTorcuco \ : required tho sum of tweuty- 

 0110 terms. 



Geometrical Progression. 



In an arithmetical progression, tho several terms give 

 equal differences ; in a geometrical progression they give 

 equal quotients ; and this constant quotient, arising 

 from dividing any term by that which immediately pre- 

 cedes it, is called tho common ratio. Tho general form 

 for a geometrical scries, continued as far as n terms, is 



a, ar, or 2 , or 3 , or*, or"" 1 



where a is the first term, and r the common ratio ; the 

 last, or nth term, being ( = or"" 1 ; which formula en- 

 ables us readily to find any remote term (tho nth term) 

 when the first term and the common ratio are given. 



By glancing at the above general form for a geo- 

 metrical series, you will at once see that the product of 

 any two terms is always equal to the product of any 

 other two equidistant from thorn ; and that if tho 

 number of terms be odd, tho product of the < rtreme I 

 terms is always equal to the square of tho middle or 

 mean term. Thus : the product of tho first and fifth 

 terms is equal to that of the second and fourth, as also 

 to the square of tho third : that is 



ar 1 X a = ar 3 X ar = (ar-)- = a 2 r*. 



When the first term of a geometrical progression is 

 given, it is plain that we only require to know the 

 common ratio in order to write down the succeeding term:) 

 to any extent ; and from this it follows that when tho 

 leading term and any remote term are given, wo may 

 always find the intervening terms, and thus fill up tho 

 gap : for instance, if a and ar* be given, to find tho tiiree 

 intervening terms, we should divide or 4 by a; and knowing 

 from the fact that there are /ice terms altogether that 

 tho quotient would be tho fourth power of the ratio, we 

 should obtain the ratio itself by taking tho fourth root of 

 that quotient : by help of the ratio, tho wanting terms 

 may be easily supplied, as noticed above. These in- 

 tervening terms are called geometrical means between tho 

 given extremes ; if only one mean is to be inserted 

 between two extremes, it is found by taking tho square 

 root of tho product of tho extremes. Suppose, fur 

 example, it were required to find a geometrical mean 

 between 2 and 8 : then, since the square root of 8 X 2, 

 or 16, is 4, wo know tho mean to bo this number ; tho 

 progression being 2, 4, 8 : if the proposed extremes had 

 been 2, 8, wo should have taken tho miniu root of 

 16, and have written the progression thus, 2, 4, 8 : 

 nevertheless, whether the given extremes bo both plus or 

 both HUNK.';, tho square root of their product with > 

 sign may be truly regarded as a menu, for the following 

 arc geometrical progressions, as well as those above, 2, 

 4, 8 ; 2, 4, 8 : tho common ratio here being 2 ; 

 in the sets above it was + 2. 



Again, suppose we had to insert two geometrical means 

 between 3 and 81 : then, as there are four terms alto- 

 gether, tho exponent n 1 above, is hero 3, /. 81 -s- 3 

 = 27=r 3 .'. r = 3. Ilenco tho progression is :;, '.'. L'7, 

 81; tho required means being 9 and LY. 



To find the sum (S) of n terms of a geometrical progressing 



Since 8 = o + ar + or 2 + or' + or 4 + . . . or"- 2 



+ ar--' 

 . . Sr = ar + ar 2 +ar s + ar*+ . . . ar"- 2 



+ ar" ' + ar". 

 Subtracting tho first expression from the second 



Sr-S + ar" a . -. S - ?^~, or - a!^! 1 . 



f* ^" X f *^* 1 . 



