for a complete divisor 1261, which, being multiplied by the quotient 1 and subtracted from the dividend, leaves a difference of 816.

Annexing the next period to this difference, we have for a dividend 816696.

Annexing a cipher to the root 21, before found, squaring the result, and multiplying it by 3, we have for a trial divisor 132300, which is contained in the dividend about 6 times. Multiplying the root before found, with a cipher annexed, 210, by the root 6 just found, and this product by 3, we have 3780. Squaring the root just found, we have 36; adding these partial divisors, we have for a complete divisor 136116, which being multiplied by the quotient 6 produces 816696, and this subtracted from the dividend leaves no remainder. The cube root of 10077696 is therefore 216.

From the foregoing principles and the explanation we have the following rules for Cube Root:

RULES.

1. Separate the number into periods of three figures each, beginning with the units' place.

2. Find the greatest cube contained in the first period on the left; this is the first figure of the root. Subtract the cube of this root from the first period on the left, and to the remainder annex the next period to form a dividend.

3. Consider the root just found as tens, square it, and multiply by 3; this forms a trial divisor. Divide the dividend by this divisor, and the quotient forms the second figure of the root.

4. To complete the divisor, multiply the first figure of the root, with a cipher annexed, by the second figure, and this product by 3; also take the square of the second figure, add these two results to the trial divisor, and the divisor becomes complete.

5. Multiply the complete divisor by the second figure of the root, and subtract the product from the dividend.

6. If there be a remainder and any more periods, to this remainder annex the next period, and proceed as before, by annexing a cipher to the root already found, squaring it, and multiplying by 3 to find a trial divisor; this, divided into the dividend, will give the next figure of the root.

7. Complete the divisor as before, multiply by the last figure

of the root found, and subtract the product from the dividend. If there be no remainder, the number found is the cube root of the given number.

NOTE 1.—If after using up all the periods there be a remainder, the number is not a perfect cube, and its root may be further found by annexing ciphers and finding its cube root to one or more decimal places.

2. In extracting the cube root of decimals, point off from left to right, beginning at the decimal point.

3. Should the dividend at any time not contain the trial divisor, place the cipher in the root and annex another cipher to the previous root before squaring and multiplying by 3 to find a trial divisor.

1. What is the inside measure of a cubic box which shall

hold just a bushel?

Ans. 12.89+in.

2. What is the height of a cubical grain-bin that holds just 200 bu.?

Ans. 75.49-in.

3. What is the difference between 6 cu. in. and 6 in. cube? Ans. 210 cu. in.

4. A cubic pile of wood contains 80 cd.: what are its dimensions? Ans. 21.71+ft. cube.

5. What is the size of a cubic cistern which shall hold 1500 gal. of water? Ans. 70.24-in. cube.

6. A man has a hall which he wishes to paint, including the floor; its measure is 15 ft. high, 36 ft. wide and 50 ft. long: how large must a cubic hall be to contain the same number of cubic feet of air; and which is the more expensive to paint, and how much at 5 cts. per square foot?

Ans. Size of hall, 30 ft. cube; 1st cost $39 more than the 2d.

SECTION III.

SIMILAR SURFACES AND VOLUMES.

302. Similar Volumes are such as have the same shape, though they may differ in size.

According to the principles of geometry,

1. Similar surfaces are to each other as the squares of their like dimensions.

2. Similar volumes are to each other as the cubes of their like dimensions.

3. The dimensions of similar volumes are to each other as the cube roots of their contents or the square roots of their surfaces.

PROBLEMS.

1. If a half-inch pipe discharge a gallon of water in a certain time, how much will a 2-in. pipe discharge in the same time?

STATEMENT.-(1)2 : (2)2 : : 1 gal. : (?).

2. If a building-lot 40 ft. square cost $200, what will a lot 100 ft. square cost at the same rate? Ans. $1250.

3. If the area of a circle whose diameter is 10 ft. is 78.54 sq. ft., what is the area of a circle whose diameter is 50 ft.? Ans. 1963.50 sq. ft.

4. If a water-pipe 1 in. in diameter discharge 1 gal. in 40 sec., how much will a 2-ft. pipe discharge in a minute?

Ans. 864 gal.

5. A piece of ground 200 ft. square was sold for $1920: how much at the same rate is a piece 25 ft. square worth?

Ans. $30.

6. If a 5-in. cube of marble weighs 12 lb., what will a cubic Ans. 165 lb. 14,228 oz. foot weigh?

STATEMENT.-5 in.3 : 12 in.3 :: 12 lb. : (?).

26

7. If a ball of metal 2 in. in diameter weigh 5 lb., what will a ball of the same metal 6 in. in diameter weigh?

Ans. 135 lb.

8. If a man 6 ft. high weigh 180 lb., what will one 5 ft. high of the same proportions weigh? Ans. 104 lb.

9. If-inch round iron weigh 1 lb. to the foot, how much ought 21-in. round iron to weigh? Ans. 25 lb. to the foot.

10. Which weighs the most—an apple 3 in. in diameter, or 3 apples 2 in. in diameter? Ans. 1 apple 3 in. in diameter.

CHAPTER XIII.

PROGRESSIONS.

303. A Progression is a series of numbers which increase or decrease according to a fixed law.

304. Progressions are of two kinds-Arithmetical and Geometrical.

305. The Terms of a progression are the numbers of which the series consists.

306. The first and the last term are called the Extremes, and all the others the Means.

SECTION I.

ARITHMETICAL PROGRESSION.

307. In an Arithmetical Progression the terms increase or decrease by a common difference. Thus, 1, 4, 7, 10, 13, 16, etc. is an increasing arithmetical progression whose common difference is 3; and 10, 8, 6, 4, 2 is a decreasing arithmetical progression whose common difference is 2.

308. The quantities considered are—

1. The first term (a).

2. The last term (1).

3. The common difference (d). 4. The number of terms (n).

5. The sum of the terms (S.).

CASE I.

To find either Extreme when the other Extreme, the Common Difference and the Number of Terms are given.

WRITTEN PROBLEMS.

1. The first term of a series is 5, the common difference 3, and the number of terms 15: what is the last term?

And in the same manner to the 15th term, which equals the 1st+(14×3), or 5+ (14x3).

2. The last term of an arithmetical series is 40, the common difference 4, and the number of terms 8: what is the first term?

8-1-7 40-(7×4)=12

Explanation. The 8th term = a+7d; hence, a= =40 — 7d=40 — (7 × 4) = 12.

From the foregoing solutions we derive the following

RULES.

1. If the smaller extreme be given, add to it the product of the common difference and the number of terms, less one, to find the larger.

2. If the larger extreme be given, subtract from it the product of the common difference and the number of terms, less one, to find the smaller.

These rules may be expressed by the following formulas : (1.) la(n-1) d.

(2.) a =l (n-1) d.

3. The first term of an increasing series is 5, the common difference 4, and the number of terms 30: what is the last

term?

Ans. 121.