Fig. IV. shows the 'pile which 13824 solid blocks of one foot each would make, when laid together, and the root, 24, shows the length of one side of the pile. The correctness of the work may be ascertained by cubing the side now found, 243, thus, 24 x 24 x 24 = 13824, the given number; or, it may be proved by adding together the contents of all the several parts, thus,

addition to the sides a, b, and c, Fig. I.

addition to fill the deficiencies n, n, n, Fig. II.

addition to fill the corner e, e, e, Fig. IV.

contents of the whole pile Fig. IV., 24 feet on each side.

From the foregoing example and illustration, we derive the following

RULE

1. Separate the given number into periods of three figures each, by putting a point over the urit figure, and every third figure beyond the place of units.

II. Find the greatest cube in the left hand period, and put its root in the quotient.

III. Subtract the cube thus found from the said period, and to the remainder bring down the next period, and call this the dividend.

IV. Multiply the square of the quotient by 300, calling it the divisor.

V. Seek how many times the divisor may be had in the dividend, and place the result in the root; then multiply the divisor by this quotient figure, and write the product under the dividend.

VI. Multiply the square of this quotient figure by the former figure or figures of the root, and this product by 30, and place the product under the last; under all write the cube of this quotient figure, and call their amount the subtrahend.

VII. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before; and so on, till the whole is finished. Note 1. If it happens that the divisor is not contained in the dividend, a cipher must be put in the root, and the next period brought down for a dividend.

Note 2. The same rule must be observed for continuing the operation, and pointing off for decimals, as in the square root.

Note 3. The pupil will perceive that the number which we call the divisor. when multiplied by the last quotient figure, does not produce so large a number as the real subtrahend; hence, the figure in the root must frequently be smaller than the quotient figure.

EXAMPLES FOR PRACTICE.

6. What is the cube root of 1860867?

What is a cube? 2. What is understood by the cube root? 3. What is it to extract the cube root? 4. Why is the square of the quotient multiplied by 300 for a divisor? 5. Why, in finding the subtrahend, do we multiply the square of the last quotient figure by 30 times the former figure of the root? 6. Why do we cube the quotient figure? 7. How do we prove the operation?

EXERCISES.

1. What is the side of a cubical mound, equal to one 288 feet long, 216 feet broad, and 48 ft. high? Ans. 144 feet. 2. There is a cubical box, one side of which is 2 feet; how many solid feet does it contain? Ans. 8 feet. 3. How many cubic feet in one 8 times as large; and what would be the length of one side? Ans. 64 solid feet, and one side is 4 feet 4. There is a cubical box, one side of which is 5 feet; what would be the side of one containing 27 times as much? 64 times as much? 125 times as much? Ans. 15, 20 and 25 ft. 5. There is a cubical box measuring 1 foot on each side; what is the side of a box 8 times as large? 27 times?64 times? Ans. 2, 3, and 4 ft.

T 111. Hence, we see that the sides of cubes are as the cube roots of their solid contents, and, consequently, their contents are as the cubes of their sides. The same proportion is true of the similar sides, or of the diameters of all solid figures of similar forms.

6. If a ball, weighing 4 pounds, be 3 inches in diameter, what will be the diameter of a ball of the same metal weighing 32 pounds? 4:32:33: 63 Ans. 6 inches. 7. If a ball, 6 inches in diameter, weigh 32 pounds, what will be the weight of a ball 3 inches in diameter 7' Ans. 4 lbs. 8. If a globe of silver, 1 inch in diameter, be worth $6, what is the value of a globe 1 foot in diameter ? Ans. $10368. 9. There are two globes; one of them is 1 foot in diameter, and the other 40 ft. in diameter; how many of the smaller globes would it take to make 1 of the larger? Ans. 64000. 10. If the diameter of the sun is 112 times as much as the diameter of the earth, how many globes like the earth would it take to make one as large as the sun? Ans. 1404928. 11. If the planet Saturn is 1000 times as large as the earth, and the earth is 7900 miles in diameter, what is the diameter of Saturn ? Ans. 79000 miles.

12. There are two planets of equal density; the diameter of the less is to that of the farger as 2 to 9; what is the ratio of their solidities?

Ans.; or, as 8 to 729.

Note. The roots of most powers may be found by the square and cube root only; thus, the biquadrate, or 4th root, is the square root of the square root; the 6th root is the cube root of the square root: the 8th root is the square root of the 4th root; the 9th root is the cube root of the cube root, &c. Those roots, viz. the 5th, 7th, 11th, &c., which are not resolvable by the square and cabe roots, seldom occur, and when they do, the work is most easily performed by logarithms; for, if the logarithm of any number be divided by the index of the root, the quotient will be the logarithm of the root itself.

ARITHMETICAL PROGRESSION.

¶ 112. Any rank or series of numbers, more than two, increasing or decreasing by a constant difference, is called an Arithmetical Series, or Progression.

When the numbers are formed by a continual addition of the common difference, they form an ascending series; but when they are formed by a continual subtraction of the common difference, they form a descending series.

3, 5, 7, 9, 11, 13, 15, &c. is an ascending series. Thus, 15. 13, 119, 7, 5, 3, &c. is a descending series.

The numbers which form the series are called the terms of the series. The first and last terms are the extremes, and the other terms are called the means.

There are five things in arithmetical progression, any three of which being given, the other two may be found:

1st. The first term.

2d. The last term.

3d. The number of terms.

4th. The common difference.

5th. The sum of all the terms.

1. A man bought 100 yards of cloth, giving 4 cents for the first yard, 7 cents for the second 10 cents for the third, and so on, with a common difference of 3 cents; what was the cost of the last yard?

As the common difference, 3, is added to every yard except the last, it is plain the last yard must be 99 3,297 cents more than the first yard. Ans. 301 cents. Hence, when the first term, the common difference, and the number of terms, are given, to find the last term,-Multiply the number of terms, less 1, by the common difference, and add the first term to the product for the last term.

2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term? Ans. 301. 3. There are, in a certain triangular field, 41 rows of corn; the first row, in I corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row ? Ans. 81 hilis.

4. A man puts out $1, at 6 per cent. simple interest, which, in 1 year, amounts to $106, in 2 years to $112, and so on, in arithmetical progression, with a common difference of $'06; what would be the amount in 40 years? Ans. $340.

Hence, we see that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the principal is the first term, the last amount is the last term, the yearly interest is the common difference, and the number of years is 1 less than the number of terms. 5. A man bought 100 yards of cloth in arithmetical progression; for the first yard he gave 4 cents, and for the last 301 cents; what was the common increase of the price on each succeeding yard?

This question is the reverse of example 1; therefore, 301-4=297, and 29799—3, common difference.

Hence, when the extremes and number of terms are given, to find the common difference,Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference.

6. If the extremes be 5 and 605, and the number of terms 151, what is the common difference? Ans. 4. 7. If a man puts out $1, at simple interest, for 40 years, and receives, at the end of the time, $340, what is the rate? If the extremes be 1 and 3'40, and the number of terms 41, what is the common difference? Ans '06. 8. A man had 8 sons, whose ages differed alike; the youngest was 10 years old, and the eldest 45; what was the common difference of their ages? Ans. 5 years.

9. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for the first yard, and 301 cents for the last yard; what was the average price per yard, and what was the amount of the whole?

Since the price of each succeeding yard increases by a constant excess, it is plain, the ave rage price is as much less than the price of the last yard, as it is greater than the price of the first yard; therefore, one half the sum of the first and last price is the average price. One half of 4 cts. 301 cts. 152 cts. average price; and the price, 1524 cts. X100 15250 cts. $15250, whole cost.

Ans. Hence, when the extremes and the number of terms are given, to find the sum of all the terms,-Multiply one half the sum of the extremes by the number of terms, and the product will be the answer.

10. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series? Ans. 46055. 11. What is the sum of the first 100 numbers, in their natural order, that is, 1, 2, 3, 4, &c. ?

Ans. 5050.

Ans. 78.

12. How many times does a common clock strike in 12 hours? 13. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent. simple interest, for the use of the money? The last year's rent will evidently be $50 without interest, the last but one will be the amount of $50 for 1 year, the last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years = $107.

If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series ? Ans. $1570. 14. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent. simple interest ? Ans. $7900.

15. There are, in a certain triangular field, 41 rows of corn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how many hills of corn in the field? Ans. 1681 hills.

16. If a triangular piece of land, 30 rods in length, be 20 rods wide at one end, and come to a point at the other, what number of square rods does it contain? Ans. 300. 17. A debt is to be discharged at 11 several payments, in arithmetical series, the first to be $5, and the last $75; what is the whole debt? the common difference between the se veral payments 7 Ans, whole debt, $140; common difference, $7.

18. What is the sum of the series 1, 3, 5, 7, 9, &c., to 1001?

Ans. 251001.

Note. By the reverse of the rule under ex. 5, the difference of the extremes 1000, divided by the common difference 2, gives a quotient which, increased by 1, is the number of terms=501. 19. What is the sum of the arithmetical series 2, 24, 3, 3, 4, 44, &c., to the 50th term inclusive? Ans. 712

20. What is the sum of the decreasing series 30, 29, 291, 29, 28, &c. down to 0? Note. 30+1=91, number of terms.

QUESTIONS.

Ans. 1365.

1. What is an arithmetical progression? 2. When is the series called ascending? 3. when descending? 4. What are the numbers forming the progression called? 5. What are the first and last terms called? 6. What are the other terms called? 7. When the first term, common difference, and number of terms, are given, how do you find the last term? 8. How may arithmetical progression be applied to simple interest? 9. When the extremes and number of terms are given, how do you find the common difference? 10. How do you find the suin of all the terms?

GEOMETRICAL PROGRESSION.

T 113. Any series of numbers, continually increasing by a constant multiplier, or decreasing by a constant divisor, is called a Geometrical Progression. Thus, 1, 2, 4, 8, 16, &c., is an increasing geometrical series, and 8, 4, 2, 1, 1, 4, &c., is a decreasing geometrical

series.

As in arithmetical, so also in geometrical progression, there are five things, any three of which being given, the other two may be found:

1st. The first term. 2d. The last term. 3d. The number of terms. 4th. The ratio. 5th. The sum of all the terms.

The ratio is the multiplier or divisor by which the series is formed.

1. A man bought a piece of silk measuring 17 yards, and by agreement was to give what the last yard would come to, reckoning 3 cents for the first yard, 6 cents for the second, and so on, doubling the price to the last: what did the piece of silk cost him?

3 X 2 X 2 X 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 196603 cents, $1966 08, Answer.

In examining the process by which the last term (196608) has been obtained, we see that it is a product, of which the ratio (2) is sixteen times a factor, that is, one time less than the number of terms. The last term, then, is the sixteenth power of the ratio, (2,) multiplied by the first term (3.)

Now, to raise 2 to the 16th power, we need not produce all the intermediate powers; for 21 =2X2 X2 X2 16, is a product of which the ratio 2 is four times a factor; now, if 16 be multiplied by 16, the product, 256, evidently contains the same factor (2) 4 times+4 times, 8 times; and 256 X 256 65536, a product of which the ratio (2) is 8 times +8 times, 16 times, factor; it is, therefore, the 16th power of 2, and, multiplied by 3, the first term, gives 196603, the last term, as before. Hence,

When the first term, ratio, and number of terms, are given, to find the last term,

I. Write down a few leading powers of the ratio with their indices over them.

II. Add together the most convenient indices, to make an index less by one than the number of the term sought.

III. Multiply together the powers belonging to those indices, and their product, multiplied by the first term, will be the term sought.

2. If the first term be 5, and the ratio 3, what is the Sth term ?

Powers of the ratio, with 1 2 3+4=

7

their indices over them. 3, 9, 27, × 81 – 21375 first term, = 10935, Answer.

3. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels: these he plants the second year; now, supposing the annual increase to continue 8 fold, what would be the produce of the 15th year, allowing 1000 kernels to a pint ? Ans. 2199023255 552 bushels. 4. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years? 217131072.

Ans. $1310 72.

5. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 cents for the second, and so on, in 3 fold ratio; what did the whole cost him? 2+6+18+5180 cts. Ans. 80 cts. In a long series, the process of adding in this manner would be tedious. Let us try, there fore, to devise some shorter method of coming to the same result. If all the terms, excepting the last, viz. 2+6+18, be multiplied by the ratio, 3, the product will be the series 6+18 +54; subtracting the former series from the latter, we have, for the remainder, 51-2, that is, the last term, less the first term, which is evidently as many times the first series (26 +18) as is expressed by the ratio, less 1; hence, if we divide the difference of the extremes (512) by the ratio, less 1, (3-1) the quotient will be the sum of all the terms, excepting the last, and, adding the last term, we shall have the whole amount. Thus, 54-2 52, and 312; then, 522≈ 26, and 51 added, makes 80, Answer, as before.

Hence, when the extremes and ratio are given, to find the sum of the series,-Divide the difference of the extremes by the ratio, lees 1, and the quotient, increased by the greater term, will be the answer.

6. If the extremes be 4 and 131072, and the ratio 8, what is the whole amount of the series? 131072-4+ 131072 ≈ 149795, Ans.

1 1

7. What is the sum of the descending series 3, 1, 3, 27, &c., extended to infinity?

It is evident the last term must become 0, or indefinitely near to nothing; therefore, the extremes are 3 and 0, and the ratio 3.

1

Ans. 44.

Ans. 1.

8. What is the value of the infinite series 1++ 16 +

1

64'
1

4, &c. 7

1

10 +100+

100 1000 +10000, &c., or, what is

9. What is the value of the infinite series the same, the decimal '11111, &c., continually repeated?

Ans.

10. What is the value of the infinite series $100 +10000, &c., descending by the ratic 100,

or, which is the same thing, the repeating decimal '020202, &c. ?

Ans..

11. A gentleman, whose daughter was married on a new year's day, gave her a dollar, promising to triple it on the first day of each month in the year; to how much did her portion amount?

Here, before finding the amount of the series, we must find the last term, as directed in the rule after ex. 1. Ans. $265'720. The two processes of finding the last term and the amount may, however, be conveniently reduced to one, thus:

When the first term, the ratio, and the number of terms, are given, to find the sum or amount of the series, Raise the ratio to a power whose index is equal to the number of terms, from which subtract 1; divide the remainder by the ratio, less 1, and the quotient, multiplied by the first term, will be the answer.

Applying this rule to the last example, 312- 531441, and

531441-1

3-1

X1 = 265720.
Ans. $265 720, as before.

12. A man agrees to serve a farmer 40 years without any other reward than 1 kernel of corn for the first year, 10 for the second year, and so on, in 10 fold ratio, till the end of the time; what will be the amount of his wages, allowing 1000 kernels to a pint, and supposing he sells

his corn for 50 cents per bushel ? 1010-1x1 = {1, 111, 111, 111, 111, 111 kernels,

Ans. $8,680,555,555,555,555,555,555,555,555,555,555,555-555

1000. 13. A gentleman, dying, left his estate to his 5 sons; to the youngest $1000, to the second $1500, and ordered that each son should exceed the younger by the ratio of 14; what was the amount of the estate ?

Note. Before finding the power of the ratio 14, it may be reduced to an improper fraction = or to a decimal, 1'5.

35
2-1
-1

1'55-1

X1000-$13187; or, 15-TX1000-$13187 50, Ans.

Compound Interest by Progression. -

114. 1. What is the amount of $4, for 5 years, at 6 per cent. compound interest? We have seen, ( 92,) that compound interest is that which arises from adding the interest to the principal at the close of each year, and, for the next year, casting the interest on that amount, and so on. The amount of $1 for 1 year is $1'06; if the principal, therefore, be multiplied by 1'06, the product will be its amount for 1 year; this amount, multiplied by 1'06, will give the amount (compound interest) for 2 years; and this second amount, multiplied by 106, will give the amount for 3 years; and so on. Hence, the several amounts arising from any sum at compound interest form a geometrical series, of which the principal is the first term; the amount of $1 or 1., &c., at the given rate per cent., is the ratio; the time, in years, is 1 less than the number of terms; and the last amount is the last term.

The last question may be resolved into this: If the first term be 4, the number of terms 6, and the ratio 1'06, what is the last term? 1'065-1338, and 1338 x 4-$5352+. Aus. $5.352. Note 1. The powers of the amounts of $1, at 5 and at 6 per cent., may be taken from the table under 91. Thus, opposite 5 years, under 6 per cent., you find 1'338, &c.

Note 2. The several processes may be conveniently exhibited by the use of letters; thus. Let P. represent the Principal.

When two or more letters are joined together, like a word, they are to be multiplied toge ther. Thus PR. implies that the principal is to be multiplied by the ratio. When one letter is placed above another, like the index of a power, the first is to be raised to a power whose index is denoted by the second. Thus RT. implies that the ratio is to be raised to a power, whose index shall be equal to the time, that is, the number of years.

2. What is the amount of 40 dollars for 11 years, at 5 per cent. compound interest? RT. XP. A.; therefore, 1'05114068'4.

ARS. $68'40.

3. What is the amount of $6 for 4 years, at 10 per cent. compound interest? Ans. $9'784 4. If the amount of a certain sum for 5 years, at 6 per cent, compound interest, be $5'352, what is that sum, or principal?

If the number of terms be 6, the ratio 1'06, and the last term 5,352, what is the first term?
A.
This question is the reverse of the last; therefore

5. What principal, at 10 per cent. compound interest, will amount, in 4 years, to $37816?

Ans. $6.

6. What is the present worth of $68 40, due 11 years hence, discounting at the rate of 5 per cent compound interest ?

Ans. $40.