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3. To the remainder bring down the first figure in the next period, and call it the dividend.

4. Involve the root to the next inferiour power to that which is given, and multiply it by the number denoting the given power, for a divisor.

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

6. Involve the whole root to the given power, and subtract it from the given number as before.'

7. 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, until the whole is finished.

EXAMPLES
1. What is the cube root of 53157376 ?

53:57376(376 root.
27=33

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53157376=3763 2. What is the biquadrate root of 34827998976?

Ans. 431,9. + 3. What is the sursolid root of 281950621875 ?

Ans. 195. 4. What is the square cubed, or sixth root of 16196, 005304479729?

Ans. 5,03. . 5. Find the seventh root of 34487717467,30751.

" Ans. 32,01.+ . NOTE.-The roots of most powers may be found by the square and cube roots only; therefore, when any even power is given, the better way will be especially in very high powers, to extract the square root of it, which reduces it to half the given power, then the square root of that power ; and so on till it comes to a square or cube.

For example, suppose a 12th power be given ; the square root of that reduces it to a sixth power; and the square root of a sixth power to a cube.

6. Extract the eighth root of 7213895789838336.

Aps. 96. 7 What is the biquadrate root of 5308416 ? Ans. 48.

ARITHMETICAL PROGRESSION.

Any rank of numbers more than two increasing by a common excess, or decreasing by a common difference, is said to be in Arithmetical Progression: such are the numbers 1, 2, 3, 4, &c. 7, 5, 3, 1; and ,8, ,6, ,4, ,2. When the numbers increase, they form an ascending series; but when they decrease, they form a descending series,

The numbers which form the series, are called the terms of the progression.

Any three of the five following terms being given, the other two may readily be found.

1st. The first term, Los 2d. The last term, je

commonly called the extremes. 3d. The number of terms. 4th. The common difference. 5th. The sum of all the terms.

PROBLEM 1. The first term, the last term, and the number of terms be: ing 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, the number of terms 11; required the sum of the series.

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2. How many strokes does a Venice clock strike in the compass of a day, going to 24 o'clock ? Ans. 300.

3. If 100 stones be placed in a right line a yard distant from each other, and the first a yard from a basket; what distance will that man go who gathers them up singly, returning with them one by one to the basket ?

Ans. 5 miles and 1300 yards. . 4. A draper sold 100 yards of cloth at 5cts. for the first yard, 10cts. for the second, 15 for the third, &c., increasing 5cts. for every yard; what did the whole amount to, and what did it average per yard ? Ans. Amount was $252), and the average price was

: $2,52cts. 5m. per yard.

PROBLEM II. The extremes and the number of terms being 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 3 and 19, and the number of terms is 9; required the common difference, and the sum of the whole series.

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2 difference. 2. A man is to travel from Boston to a certain place in 12 days, and to go but three miles the first day, increasing every day by an equal excess, so that the last day's journey may be 58 miles; required the daily increase, and the distance of the place from Boston.

. Ans. Daily increase 5, distance 366 miles. 3. A man had 12 sons whose several ages differed alike; the eldest was 49, the youngest 5 years old ; what was the common difference of their ages ? Ans. 4 years. PROBLEM III. Given the first term, the last term, and the common differ

ence, to find the number of terms. Rule.—Divide the difference of the extremes by the common difference, and the quotient increased by 1, is the number of terms required.

EXAMPLES. 1. The extremes are 3 and 19, and the common difference 2 ; what is the number of terms?

19

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Ans. 9 2. Suppose a man travel the first day 7 miles, the last 51 miles, and increase his journey each day by 4 miles ; how many days will he travel, and how far ?

Ans. 12 days, and 348 miles.

GEOMETRICAL PROGRESSION.

Any series of numbers, the terms of which gradually increase or decrease by a constant multiplication or division, are said to be in Geometrical Progression. Thus, 4, 8, 16, 32, 64, &c. and 81, 27, 9, 3, 1, &c. are series in geometrical progression, the one increasing by a constant multiplication by 2, and the other decreasing by a constant division by 3.

The number by which the series is constantly increase ed or diminished, is called the ratio.

PROBLEM I. Given the first term, the last term, and the ratio, to find

the sum of the series. RULE.—Multiply the last term by the ratio, and from the product subtract the first term, and the remainder divided by the ratio less 1, will give the sum of the series. EXAMPLES. 1. The extremes of a geometrical progression are 1 and 65536, and the ratio 4 ; what is the sum of the series?

65536

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87381 Ans. 2. A man travelled 6 days; the first day he went 4 miles, and doubling his travel each day, his last day's ride was 128 miles; how far did he go in the whole ?

Ans. 252 miles. 3. The extremes of a geometrical series are 1024 and • 59019, and the ratio is 1; what is the sum of the series?

Ans. 175099.

PROBLEM II.
Given the first term and the ratio, to find any other term

assigned.* CASE I.- When the first term of the series, and the ratio,

are equal.t. RULE.--1. Write down a few of the leading terms of the 'series, and pface their indices over them, beginning the indices with a unit or 1. .

2. Add together such indices as, in their sum, shallmake up the entire index to the term required.

• As the last term in a long series of numbers is very tedious to be found by continual multiplication, it will be necessary for more readily finding it out, to have a series of numbers in arithmetical proportion, called indices, whose common difference is 1.

+ When the first term of the series, and the ratio, are equal, the indices must begin with a unit, and in this case, the product of any two terms is equal to that term, signified by the sum of their indices.

how $ 1, 2, 3, 4, 5, &c. Indices or arithmetical series. Thus

s: 2, 4, 8, 16, 32, &c. Geometrical series. Now, 2+3=5, the index of the fifth term, and 4x8= 32=the fifth terma,

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