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is the 6th root of x6; the 4th root of x is xạ, whose square root, X, is the gth root of x8; the cube root of 29 is x?, whose cube root, x, is the 9th root of ro; the 4th root of x'2 is 2*, whose cube root, x, is the 12th root of 22 ; &c.

244. Fractional Indices.— The employment of fractional indices enables us to write roots under the form of powers--a very convenient form in many cases. Thus, instead of


NT

63
we can write ct

ci

a

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yi

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&c.

&c. Because ($ 220) the square of al is a; the cube of 7% is 6; the fourth power of ct is c; the square of că is 23 ; the cube of y& is yo; &c.: (a)=at xalat+ia=a; (61)3=8x 68 x 6

= +33+3+3=0=b; (c) '0(

a*xe*xx_++++++ ( Tỉy=x*xx=x+z=a; (*)=x*xy=

„3+3+1

yo; &c. The expressions at, 63, and ct would be read "a in the power ],"«b in the power }," and "c in the power ,” respectively. Such an expression as zi or yš—the numerator of the index not being unity-could be read in either of two ways : x is “ the square root of the cube of x,”

C," or

the cube of the square root of x;" and yg is “ the cube root of the square of y,"

square of the cube root of y.It has already been shown that xl is the square root of x®, and yế the cube root of ij; and it can be shown, quite as easily, that al is the cube of r*, and y the square of y#: (zł)=x* xr xrt=+1+=21; (ytje=yt xy

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PROGRESSION. 245. By a PROGRESSION is meant-a series of three or more numbers which successively increase, or successively decrease, at some uniform rate.

246. The numbers forming a progression are called its TERMS, of which the first and the last are

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known as the extremes, and the intermediate terms as the means.

247. A progression is said to be ARITHMETICAL (or equidifferent) when, of every three consecutive terms, the difference between the first and the second is equal to the difference between the second and the third; GEOMETRICAL (or equirational), when, of every three consecutive terms, the ratio which the first bears to the second is equal to the ratio which the second bears to the third; and HARMONICAL, when, of every three consecutive terms, the first bears to the third the same ratio which the difference between the first and the second bears to the difference between the second and the third.

248. A progression is called an ascending or a descending one--according as the terms increase or decrease from left to right.

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ARITHMETICAL PROGRESSION. 249. Being given the first term and the common difference, we can form an arithmetical progression by continually adding the common difference to, or continually subtracting it from, the first term-. according as the progression is “ascending" or

” “descending."

EXAMPLE I.-Set down the ascending arithmetical progression which has 1 for its first term, and 2 for common difference.

Adding 2 to 1, we obtain the second term, 3; adding 2 to 3, we obtain the third term, 5; adding 2 to 5, we obtain the fourth term, 7; and so on. The progression, therefore, is

3 5 7 9 II 13 15 17 19 Example II.--Set down the descending arithmetical progression which has 35 for its first term, and 3 for common difference.

Subtracting 3 from 35, we obtain the second term, 32; subtracting 3 from 32, we obtain the third term, 29; subtracting 3 from 29, we obtain the fourth term, 26; and so on. The progression, therefore, is

35 32 29 26 23 17 14 8 5 From an examination of the preceding examples it will be

I

21

23 &c.

20

II

2

seen that, in the case of an ascending arithmetical progression, the second term exceeds the first by the common difference; that the third term exceeds the first by twice the common difference; that the fourth term exceeds the first by 3 times the common difference; and so on : and that, in the case of a descending arithmetical progression, the first term exceeds the second by the common difference; that the first term exceeds the third by twice the common difference; that the first term exceeds the fourth by 3 times the common difference ; and so on.*

250. To find any term (after the first) of an arithmetical progression : Subtract 1 from the number indicating the place of the term; multiply the remainder by the common difference; and add the product to, or take it from, the first term-according as the progression is an ascending or a descending one.

EXAMPLE III.--The first term of an ascending arithmetical progression is 10, and the common difference 7; what is the 13th term ?

The 13th term exceeds the first term by 12 times the common difference. Adding 12x7, therefore, to 10, we find the required term to be 94:

13-1=12 ; 12x7=84; 10+84594. ExamPLE IV.—The first term of a descending arithmetical progression is 158, and the common difference 9; what is the i Ith term ?

The rith term is less than the first term by 10 times the common difference. Subtracting 10x9, therefore, from 158, we find the required term to be 68:

11-1=10; 10×9=90; 158-90=68. From what we have just seen, it is evident that the difference between the extremes of an arithmetical progression--whether the progression be an ascending or a descending one is the product of two factors, namely : (a) the common difference, and (6) what remains when the number of terms is diminished by 1.

251. To find the number of terms in an arith* Putting f for the first term, and d for the common difference, we have, in the case of -an ascending progression,

a descending progression,

Ist term = f
2nd
= fta

2nd =f-a
3rd =p+2d

3rd =f-2d 4th $+3d

4th =.f-3d nth = f+(n-1)d

nth f-(n-1)

Ist term

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metical progression—the extremes and the common difference being given : Divide the common difference into the difference between the extremes, and add 1 to the quotient.

ExamPLE V.—The extremes of an arithmetical progression are 7 and 43, and the common difference is 4; find the number of terms.

The difference between the extremes is 36, in which the common difference (4) is contained 9 times. The number of terms, therefore, is (9ti=)10—the terms being more numerous, by 1, than the common differences :

43-7=36; 36-459; 9+1=10. 252. To find the common difference in an arithmetical progression—the extremes and the number of terms being given : Subtract 1 from the number of terms, and divide the remainder into the difference between the extremes.

EXAMPLE VI.—The extremes of an arithmetical progression are 26 and 2, and the number of terms is 9; find the common difference.

The number of terms being 9, the number of common differences is (9-1=)8; and the sum of the common differences is (26-2=)24. Dividing 24 by 8, therefore, we find the required common difference to be 3:

26—2=24; 9-1=8; 248=3. 253. To determine the means in an arithmetical progression the extremes and the number of means being given : Find the common difference (3252), remembering that the number of terms is greater by 2 than the number of means; and then proceed in the manner already explained ($ 249).

EXAMPLE VII.—The extremes of an arithmetical progression are 11 and 39, and the number of means is 6; find the

means.

The number of means being 6, the number of terms is (6+2=8. The common difference, therefore ($ 252), is 39-11_28

= 4; so that ($249) the required means are 8-1

7 (11+4=) 15, (15+4=) 19, (19 +4=) 23, (23+4=) 27, (27+4=)31, and (31+4=)35. The progression in full is

15 19 23 27 31 35 39

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EXAMPLE VIII.-Insert 4 arithmetical means between 37

and 7.

The number of means being 4, the number of terms is (4+2 =) 6. The common difference, therefore ($ 252), is 37-7

30

Y=6; so that ($ 249) the required means are 6--1 5 (37–6=) 31, (31-6=) 25, (35-63)19, and (19-6=)13. The progression in full is

37 31 25 19 13 7 EXAMPLE IX.-Insert one arithmetical mean between 13 and 27 There being only one mean, the number of terms is (1+2=)3.

27 – 13_14. The common difference, therefore, is

3-1 the required mean is (13+7=)20.

EXAMPLE X.-Insert one arithmetical mean between 23

=7; so that

2

and 15.

As in the last case, the number of terms is (1+2=)3; so that the common difference is

23–15_8

-4.

The required

3-1 mean, therefore, is (23–4=)19.

One arithmetical mean, however, can be inserted more easily.

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254. To insert ONE arithmetical mean between two numbers: Add the numbers together, and take half their sum.

Thus, the arithmetical mean between 13 and 27 (Ex. IX.) is 13+27_40

==20; and the arithmetical mean between 23 and

2

2

15 (Ex. X.) is 23+15_38

2

2

=19. In order to understand this, let us put x for the arithmetical mean between 13 and 27, and suppose that a boy has 13 marbles in his left-hand pocket, and 27 in his right-hand pocket. Now, as 13, x, and 27 form an ascending arithmetical progression (x being greater than 13 by as much as 27 is greater than x), it is evident that, by transferring from his right-hand to his left-hand pocket a number of marbles equal to the common difference, the boy would have marbles in each pocket. We thus find x+x=13+27; 2x=13+27 ; x=

13+27 Next, let us put y for the arithmetical mean between 23 and 15, and suppose that a boy has 23 marbles in his left-hand

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