The cube of the tens' figure is thousands, and will, the greatest cube to be found in the left hand period. and its root, 20. This cube (the figure H) being 5,824 remains, (=the figures, A, B, C, D, E, Fand G.) E, F and G, are each, on one side, as long as H, (that is, may all be placed side by side, in one solid, thus,

of course, be This is 8,000, taken away, As B, C, D, 20 feet,) they

The little solid, A, must, for the present, stand by itself. If this solid, abcdef. were divided by its whole upper surface, we should obtain its thickness, a b, which is the units' figure of the root. The upper surfaces of G, F and D, we know, because each surface is equal to 20×20, or the square of the root already found. Neglecting E, C, B, and also the little cube A, then, 20×20×3=1,200, will be sufficiently near for a divisor. 5,824÷1,200-4, (neglecting remainder.) We suppose, then, that 4 is the units' figure, or the thickness, a b, which is equal to the side of A. This 4, we multiply by 202 for each of the solids G, F, D; we then square 4, and multiply it by 20, for cach of the solids, E, C, B; and finally cube it, for the solid, A. Then 202 X4×3+43×20×3+43=5,824. This agrees with what we know to be the solidity of A, B, C, D, E, F and G; and hence, we know 4 to be the units' figure of the root. Therefore, 24 is the side of a cube, containing 13,824 solid feet. The above operations show the process here illustrated. The large cube, 8,000, is first subtracted, and the remainder, 5,824, is then divided by 1,200, which is 3 times the square of 20, (the root already found.) We then multiply 3 times 202, or 1,200, by the quotient 4, 3 times 42 by 20, and, finally, cube 4. These three , results, added, form the subtrahend.

In the second operation above, the cyphers are omitted. It is evident, however, that when a divisor is obtained from the root already found, a cypher must be understood after that root. This, by squaring, will bring two cyphers at the right of the divisor. For the same reason, when a subtrahend is obtained as above directed, the first of the products which compose it, will have two cyphers, the second, one, and the third, none at the right. All these cyphers may be omitted, if they are understood, as in operation 2d, and the

numbers arranged for addition, exactly as though the cyphers were expressed. Perhaps it will be better for pupils, at first, to write them out in full, as in operation 1st. When there are more than two periods, the operation is similar. A single period is all that need be brought down at once. Periods of cyphers, (of three places each,) may be annexed at the right, if necessary, and the root car. ried to decimals. In like manner, significant decimals may be poin. ted off towards the right, from the separatrix.

It may sometimes happen that the subtrahend found as above, will be larger than the dividend. This occurs, because the divisor is smaller than the whole surface of the solid, abcdef. When this is the case, the quotient, or figure of the root last found, must be diminished, and a new subtrahend found. When no subtrahend can be obtained, smaller than the dividend, a cypher must be placed in the root, and another period brought down. 2. Extract the cube root of 48,228,544.

Cyphers retained.

Cyphers

48228544(364 Root.

27

302 X3=2700)21228 Dividend.

302 X6X3=16200 +30X62 X3 3240

+63

216

19656 Subtrahend.

362 X3=3888)1572544 Dividend.
362X4X3=15552

omitted. +36X42X3= 1728
+-43= 64

}

1572544 Subtrahend.

00

3. In making an excavation, there were thrown out 616,295,051 solid feet of earth. If it were all formed into a cubic mass, what would be the length of one side? A. 851 ft.

4. A box in the form of a cube, contains 9,261 cubic inches. What is the length of one side? A. 1 ft. 9 in.

5. From a cubical cellar were thrown out 510,082,399 ft. of earth. What was one side of the cellar? A. 799. 6. What is the side of a cubical solid, containing 988,047,936 cubic feet? A. 996 ft. 7. Find the cube root of 941,192,000. A. 980. 958,585,256. A. 986. Of 478,211,768. A. 782. 494,913,671. A. 791. Of 445,943,744. A. 764. 196,122,941. A. 581. Of 204,336,469. A. 589. 57,512,456. A. 386.

Of

Of

Of

Of

Of 6,751,269. A. 189. Of 39,651,921. A. 341. Of 42,508,549. A. 349. Of 510,082,399.

A. 799. Of 469,097,433. A. 777.

Of 41. A.

8. Find the cube root of 7. A. 1.912933. 3.448217. Of 49. A. 3.659306. Of 94. A. 4.546836.

Of 97. A. 4.610436. Of 199. A. 5.838272. Of 179. A. 5.635741. Of 389. A. 7.299893. Of 364. A. 7.140037. Of 499. A. 7.931710. Of 699. A. 8.874809. Of 686. A. 8.819447. Of 886. A. 9.604569. Of 981. A. 9.936261. Proceed with fractions as in extracting the square root. 9. Find the cube root of. A. 3. Of 458933. A. 77. Of 7301884. A. 184.

A. 273 397

26730899

Of

29797

31

1324. A. 34. 29 8 4 8 4 7 7.

0346417

10 17 3.

Of

Of. A.

10. Find the cube root of §. A. .8549879. .5593445. Of. A. .4578857.

Of 12. A. .9973262.

Of. A. .4562903.

200

From these illustrations and examples, we derive the rule,

I. HAVING Pointed off, SUBTRACT FROM THE HIGHEST PERIOD, THE GREATEST CUBE CONTAINED IN IT, PLACE THE ROOT IN THE QUOTIENT, AND, TO THE REMAINDER, BRING DOWN THE NEXT PERIOD FOR A DIVIDEND.

II. SQUARE THE ROOT ALREADY found, (UnderRSTANDING A CYPHER AT THE RIGHT,) AND MULTIPLY IT BY 3 FOR A DIVISOR. DIVIDE THE

DIVIDEND BY THE DIVISOR FOR THE NEXT FIGURE OF THE ROOT.

III. MULTIPLY THE DIVISOR BY THE QUOTIENT, MULTIPLY 3 TIMES THE SQUARE OF THE QUOTIENT BY THE PART OF THE ROOT PREVIOUSLY FOUND, FINALLY, CUBE THE QUOTIENT, ANd add these THREE RESULTS TOGETHER FOR A SUBTRAHEND.

IV. SUBTRACT THE SUBTRAHEND FROM THE DIVIDEND, TO THE REMAINDER, BRING DOWN THE NEXT PERIOD FOR A NEW DIVIDEND, AND SO PROCEED.

NOTE. The pupil's attention should be particularly called to the effect of understanding a cypher in II and III. If he does not fully understand this, he will be liable to fall into error. We have preferred the above rule, to the one usually given, because it keeps the principles of the operation before the mind, and, indeed, obliges the pupil to act upon them at almost every step. Perhaps it is not as easy as the other, in the beginning, but it will be found far more useful, and in fact, easier in the end.

The proof is by Involution.

ARITHMETICAL PROGRESSION.

§ CII. The natural series of numbers, 1, 2, 3, 4, 5, &c., consists of numbers increasing by a continual addition of 1 to the number preceding. The series, 1, 3, 5, 7, 9, &c., increases by a continual addition of 2. The series, 15, 12, 9, 6, &c., decreases by a continual subtraction of 3. Any rank, or series of numbers, consisting of more than two terms, increasing or decreasing, like the above, by a common difference, is called an ARITHMETICAL SERIES or PROGRESSION. The numbers forming the series, are called TERMS. The first and last terms, are called the EXTREMES, the others, the MEANS.

When the series increases, or is formed by a continual addition of the common difference, it is called an ASCENDING SERIES. When it decreases, or is formed by a continual subtraction, of the common difference, it is called a DESCENDING SERIES.

1. How many strokes do the clocks in Venice,, which go from 1 to 24 o'clock, strike in the course of a day?

The answer might be found by addition. But by attending to the following, an easier mode may be discovered. Suppose another clock made to strike downward from 24 to 1. The two clocks would strike equal numbers of blows in a day. The order of their striking would be as follows:

1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12, 13, 14, 15, 16, 17,18,19,20,21,22,23,24 24,23,22,21,20,19, 18, 17,16,15,14,13,12,11,10, 9, 8, 7, 6, 5, 4, 3, 2, 1 25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25 Thus, at every hour, the two clocks together would strike the same number of blows, viz. 25, and, in the course of the day, both would strike 24X25-600 blows. One would strike half this number= 300 Ans. It will be seen that 25-24+1 the sum of the extremes. Hence, the extremes and number of terms being given, to find the sum of all the terms,

MULTIPLY THE SUM OF THE EXTREMES BY THE NUMBER OF TERMS, AND HALF THE PRODUCT WILL BE THE ANSWER.

2. The first term of a series is 1, the last term 29, and the number of terms 14. What is the sum of the series? A. 210.

3. 1st. term, 2, last term, 51, number of terms, 18. Required the sum of the series, A. 477.

4. Find the sum of the natural terms 1, 2, 3, &c. to 10,000. Ans. 50,005,000.

5. A man travelled from Hartford, going 3 miles the first day, and increasing each day by an equal excess. His 12th day's journey was 58 miles. What was the daily increase, and the distance he travelled from Hartford?

As he travelled 12 days, he increased his journey 11 times, by an equal addition. 58, then, is 11 times the daily increase, more than 3. Therefore 58-3-55÷÷-11-5 daily increase. By last rule, 366 mls. distance from H,

Hence, the extremes and number of terms being given, to find the common difference,

DIVIDE THE DIFFERENCE OF THE EXTREMES BY THE NUMBER OF TERMS LESS 1.

6. Extremes 3 and 19; number of terms 9. dif.

dif.

7. Extremes 4 and 56; number of terms 14.

Required the com.
A. 2.

Required the com.
A. 4.

8. A man had 15 houses, increasing equally in value, from the first, worth $700, to the 15th worth $3,500. What was the diffe. rence in value between the first and second ? Ans. 200.

9. The ages of 5 persons were in arithmetical progression, the youngest being 15 yrs. old, and the com. dif. 2. What was the eldest

one's age? It was evidently 4 times the com. dif. more than that of the youngest. Hence, 15+4x2=23 Ans.

Hence, the least term, the number of terms, and the common difference being given, to find the greatest term,

TO THE LEAST term, ADD THE COMMON DIFFERENCE, MULTIPLIED BY THE NUMBER OF TERMS LESS 1.

4. 100 stones are a yard apart in a straight line, and the first one a yard from a basket. How far must a man go, to gather them, one by one, and return with them singly to the basket?

Ans. 5 mls. 5 fur. 36 rds. 2 yds.

NOTE. The cases of Arithmetical Progression are numerous, and of too little practical utility to warrant us in devoting to them much space. Any three of the following terms being given, the other two may be found. 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms.

GEOMETRICAL PROGRESSION.

§ CIII. The series 1, 3, 9, 27, 81, &c, consists of numbers, each of which is 3 times the preceding. The series 64, 32, 16, 8, 4, &c., consists of numbers, each of which is half the preceding. Any rank or series of numbers, of more than two terms, increasing, like the above, by a common multiplier, or decreasing by a common divisor, is called a GEOMETRICAL SERIES, or PROGRESSION.

The common multiplier or divisor is called the RATIO. The numbers which form the series are called TERMS.

The distinction between increasing or ascending, and decreasing or descending series, is made as in Arithmetical Progression.

Any three of the five following terms being given, the other two may be found. 1. The first term. 2. The last term. 3. The num. ber of terms. 4. The ratio. 5. The sum of all the terms.

1. A man bought 5 sheep, giving $1 for the first; $3 for the sec. ond; $9 for the third, and so on, in geometrical progression. What did he give for the whole ?

Write the whole series, thus, 1, 3, 9, 27, 81
Multiply by the ratio, 3, thus, 3, 9, 27, 81, 243

We here obtain a new series, having 4 terms like the last. Let the first series be subtracted from the second, and all the similar terms vanish, leaving 1, the 1st. term of the 1st. series, to be subtracted from 243, the last term of the last. 243-1-242. Now as the last series was 3 times the first, and we have subtracted once the first, the remainder, 242, must be twice the first series. HENCE, 242-2121-sum of 1st. series, Ans.

The same result would have been obtained, if we had multiplied only the last term, and taken the first from the product, since the intermediate terms vanish, in the subtraction. Hence, the rule, the extremes and ratio being given, to find the sum of all the

terms.