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6. Subtract the subtrahend from the dividend, and to the re: mainder bring down the next period for a new dividend, with which proceed as before; and so on till the whole is finished.

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Solids of the same form are in proportion to one another as the cubes of their similar sides or diameters.

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4. If a cable 12 inches round 6. There is a cistern which require an anchor of 18 cwt. what contains 8204 solid inches, I demust be the weight of an anchor | mand the side of a cubical box for a 15 inch cable ? which shall contain the same quantity. Ans. 14.12+ in.

cwt.

cwt. qr. lb. 123:18:: 153: 35017 Ans.

5. The diameter of a legal Winchester bushel is 181⁄2 inches, and its depth 8 inches; what must the diameter of that bushel be whose depth is 71⁄2 inches? Ans. 19.10671.

7. A person wanted a cylindrick vessel of 3 feet deep, that ❘ shall hold twice as much as another of 23 inches deep, and 46 | inches in diameter; what must be the diameter of the required vessel ? Ans. 57.37 in.

Between two given numbers to find two mean proportionals.

RULE. Divide the greater by the less, and extract the cube root of the quotient. Multiply the least given number by the root for the lesser, and this product by the same root for the greater of the two numbers sought.

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1. Prepare the given number for extraction by pointing off from the place of units according the required root.

2. Find the first figure of the root by trial, subtract its power from the first period, and to the remainder bring down the first figure in the next period, and call these the dividend.

3. Involve the root already found to the next inferior power to that which is given, and multiply it by the number denoting the given power, for a divisor.

4. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root.

5. Involve the whole root to the given power; subtract it from the given number as before, bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, and so on till the whole is finished.

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1. Why do you multiply the square of | 4. Why do you multiply the triple

the quotient by 300?

quotient by the square of the last

2. Why the quotient by 30?

quotient figure ?

triple square by the last quotient

of the last quotient figure ?

fpere?

What is found by multiplying the 5. Why do you add to these the cube

(

SECTION II.

Arithmetical Progression.

A rank of numbers is in Arithmetical Progression, when they increase by common excess, or decrease by a common difference. When the numbers increase, they form an ascending series, and when they decrease, a descending series. Thus, 1, 2, 3, 4, &c. and 3, 6, 9, 12, &c. are ascending series, and 10,9,8,7, &c. and 20, 16, 12, 8, &c. descending series.

The terms of the progression are the numbers which form the series. The first term and last term are called the extreme.

If any three of the five following things be given, the other two are readily found, viz. the first term, the last term, the number of terms, the common difference, and the sum of all the terms.

Problem I.

The first term, the last term, and the number of terms given, to find the sum of all the terms.

RULE.*-Multiply the sum of the extremes by the number of terms, and half the product will be the answer. Examples.

1. The first term of an arithmetical progression is 1, the last term 21, and the number of terms 11; what is the sum of the series?

21 last term.
1 first term.

-

22

11 number of terms.

22 22

2)242

121 Ans.

2. How many times does a common clock strike in 12 hours? Ans. 78 times.

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* Suppose another series of the same kind with the given one, to be placed under it in an inverse order; then will the sum of every two corresponding terms be the same as that of the first and last; consequently, any one of these sums, multiplied by the number of terms, will give the whole sum of the two series, and half this sum will evidently be the sum of the given series; thus,

24 6 8 10 given series.

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2 the same inverted.

12+12+12+12+12=12×5=60 and 6-30-2+4+6+8+10

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Problem II.

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

RULE.*-Divide the difference of the extremes by the number of terms, less 1, and the quotient will be the common difference.

Examples.

1. The extremes are 2 and 2. A man has 12 sons, whose 53, and the number of terms 18; ages are in arithmetical progreswhat is the common difference? (sion, the youngest is 2 years old, and the oldest 35; what is the common difference in their ages? Ans. 3 years.

18

1

17

53
2

17)51(3 Ans.

51

Problem III.

The first term, the last term, and common difference given, to find the number

of terms.

RULE.t--Divide the difference of the extremes by the common difference, and the quotient, increased by 1, is the number of terms required.

Examples.

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2. A man on journey, travelled the first day 5 miles; the last day 35 miles, and increased his travel each day by 3 miles; how many days did he travel? Ans. 11 days.

QUESTIONS.

L. When is a rank of numbers in | 5. What are the terms of a progres

Arithmetical Progression?

sion ?

2. What is meant by an ascending 6. What is the first problem?

series?

7. What the rule?

3. What by a descending?

8. What the second? &c.

4. What are the extremes ?

* The difference of the first and last terms evidently shows the increase of the first term, by all the subsequent additions, till it becomes equal to the last; and as the number of those additions is evidently one less than the number of terms, and the increase by every addition equal, it is plain that the total in crease, divided by the number of additions, will give the difference at every one separately; whence the rule is manifest.

+ By Problem II. the difference of the extremes, divided by the number of terms, less 1, gives the common difference; consequently, the same divided by the common difference, must give the number of terms less 1; hence this quofrent, increased the question,

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