tens, increased by 3 times the tens multiplied by the units, and the square of the units, all multiplied by the units (97); plus the cube of the units; that is, 203 + (3 × 202 × 8) + (3 × 20 × 82) +83=203+ (3 X 20+3X 20 X 8+82) X 8+83.

145. We will now extract the cube root of 21952.

- 1. The root will contain two figures,

8

tens and units. (Why?) 2. The cube t. d. 1200 21952(29

540

8

13952

16389

13952

480

13952

64

c. d. 1821
t. d. 1200

c. d. 1744

of the tens will be found in the thousands. (Why?) The greatest cube in 21 is 8, and its root, 2, will be the tens of the root. 3. Subtracting the cube of the tens from the given cube, the remainder contains three times the square of the tens, plus three times the tens multiplied by the units, plus the square of the units, all multiplied by the units. (Why?) (97.) Therefore, dividing the remainder by the former factor, viz., 3 times the square of the tens, &c., would exactly give the latter factor, viz., units. But as the units' figure is not yet found, we will take three times the square of the tens, or 12 hundreds, as a trial divisor. If we divide the remainder by this trial divisor, the quotient would be more than 11. We know the units' figure cannot be more than 9; therefore the quotient is too large, as we might expect it would be, since our divisor is but a partial divisor. A complete divisor is to be formed by adding to the trial divisor 3 times the tens multiplied by the units, and the square of the units. 3 times 2 tens multiplied by 9, or, (which is the same,) 30 times 2 multiplied by 9, is 540, and 92 is 81. Adding these to the trial divisor for a complete divisor, and multiplying by 9, we get too large a product. Our quotient figure is therefore still too large. We must erase it, and go over this part of the work again. Taking 8 as a quotient figure, we make the complete divisor 1744, which being multiplied by 8, gives the product 13952. Hence the

RULE FOR EXTRACTING THE CUBE ROOT.

1. Separate the given number into periods of three figures each, by placing a dot over every third figure, beginning at units; thus, 31486.100840. (What will the dots show? Why?)

2. Find by trial the greatest cube number in the left hand period, (which may consist of one, two, or three figures,) and place its root on the right. Subtract the cube of this root from the first period, and to the remainder bring down the 2d period for a dividend.

3. Take 300 times the square of the root already found for a trial divisor. Divide the dividend by this trial divisor, and place the result in the root. (If the trial divisor is not contained in the dividend, annex

one naught to the root, and two naughts to the trial divisor for the next trial divisor, and bring down the next period for a dividend.)

4. Take 30 times the root previously found, multiplied by the figure last placed in the root, and the square of this last figure, and add them to the trial divisor, for a complete divisor. Multiply the complete divisor by the last figure in the root; subtract the product from the dividend ; bring down the next period, and proceed as before. See 3.

The remarks concerning fractions and decimals, annexed to the rule for the extraction of the square root, are equally applicable to the cube root, by substituting the word cube for square.

The number of decimals in the cube must be a multiple of 3; as 3, 6, 9, &c.

2. Extract the cube root of 79507; 132651; 704969. (3.) 1124864; 28652616; 40001688.

(4.) 68417929; 480048687; 527514112.

5.* How much is 809? /300? /4.10846? (6.)/341?$54.0061071?

For the application of the cube root, see Art. 178 and 179.

QUESTIONS. What is the extraction of the cube root? Repeat the first 12 integral numbers, with their cubes. How many figures are

*Extract the root to the nearest ten thousandth.

there in the cube, compared with the number in the root? How many decimals? Give examples to show the truth of this. To what is the cube of a number, consisting of units and tens, equal? Show this by an example. How may the number of figures in the root be ascertained? Why? Repeat the rule for the extraction of the cube root, and give the reason for each step in the process.

SECTION XVI. — ARITHMETICAL SYMBOLS.

146. All the arithmetical symbols or signs usually employed have been introduced in different parts of this work. As the different processes in arithmetic can be so clearly and concisely indicated by means of them, it is important that the pupil should become so familiar with them, as to be able to use them properly, and to understand them when correctly used by others.

The expression 3+4x5-2 is read 3 plus 4 times 5 minus 2; and is the same as 3+20-2, or 3+18=21. 3+4×5-2, or (3+4) ×5-2, is the same as 7×5 -2-35-2 33.

The expression 5 × 4×8÷2+2 is 20 × 4+2; or, 160 ÷2+2=80+2=82. 5×4x8÷2+2 is 160 ÷ 4

40.

The expression 6 x 15÷3 × 2 is 90615; 6 X 15 3 X 2-4 is 906-415-4=11; 6 × 15÷÷ 3 X 2-4 is 90245.

The expression 3+8÷4-2 is 3+2-2-3; but 3+84-2 is 11÷4-2-2-2=}; and (38)÷ (4-2) is 112=5}.

The expression (5—3+4) × (5—1+2) ÷ (4+7) is 6×6÷11 ff=33.

The expression (1+1)ק÷(4 — }) is § × § ÷J={. X & X=3=138.

The expression (25—16) × (189) is 9/27 =3x3=9. So/29 +20-64 × is

49/64

X=7—4׆=7—1—6}; but /29+20-64 X is 49-8-7-2-5.

The expression [(§ —1⁄21⁄2÷1+8) × 4] ÷ } is [(24÷§)×

(1.) 2513—1+7. (2.) 25 + 13 — (1+7).

(3.) 185-3 x 4. (4.) 18 X 5-3 x 4.

(5.) 459-4+3x8.

(6.) 45÷9-4+3x8. (7.) 45÷9—4+3×8. (8.) (75-5) X8+4÷2+6x8-3. (9.) 75-(5x8+4)÷2+6x8-3. (10.) 75-5X (8+4÷2)+6x8-3. × × (11.) 75 ; 15 — 10+8X4−15:3+2 (12.) 75 ÷ 15—10+ (8 × 4) — 15÷3+2. (13.) §+11⁄2 −12. (14.) (§ × § + §) × (§ — §). (15.) × (+

SECTION XVII.

PROGRESSION, OR SERIES.

147. ARITHMETICAL PROGRESSION.

Arithmetical Progression, or Series by Difference, is a series of numbers which uniformly increase or decrease, so that the difference between any two adjacent numbers is the same throughout the series. This difference is called the common difference. When the common difference is added to each term to make the succeeding one, it is called an ascending series; as, 1, 3, 5, 7, &c. When it is subtracted, it is called a descending series; as, 7, 5, 3, 1.

The numbers that compose the series are called the terms of the series. The first and last terms are called the extremes, and the others, the means.

In every arithmetical series there are 5 things to be considered, viz. :

1. The first term.

2. The last term.

3. The common difference.

4. The number of terms.

5. The sum of all the terms; any three of which being given, the rest may be found.

1. If in a series the first term is 2, and the common difference 3, what is the 2d term of the series? the 3d? the 5th? the 10th term?

From this example we see that, to find the second term, we add the common difference once to the first term; to get the third term, we add the common difference twice to the first term, &c. Hence the following rules under CASE I. and II.

CASE I. The first term, common difference, and number of terms being given, to find the last term.

RULE. Multiply the common difference by one less than the number of terms, and add the product to the first term.

2. If in a series the first term is 3, and the common difference 5, what is the 4th term? the 8th? 15th? 20th ?

3. A stone falls 16.1 feet during the first second of its descent in free space, 48.3 ft. the 2d second, 80.5 ft. the 3d, &c.; how far would it fall the 8th second?

4. A man lends $100 at 6 per cent. simple interest; what will it amount to in 1 year? in 2 years? in 5 years? in 8 years? The principal and the amounts for the successive years form an arithmetical series; what is the first term? the common difference? the last term?

CASE II. The two extremes and the number of terms being given, to find the common difference.

RULE. Subtract the less extreme from the greater, and divide the remainder by 1 less than the number of terms; the quotient will be the common difference.