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Here it appears that the fifth root is between 7*3 and 74. 7*3 being taken, its fifth power is 20730*71593; Hence then

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104263.748=1 the first sum. 10426304 : 305.084 : : 73 : '0213605

73

915252 2135588

3042637)2227*1132(0213604 = difference.

208527 73

.

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3. What is the sixth root of 210358 ?

Ans. 5'254037 4. What is the seventh root of 21035:8 ?

Ans. 4*145392

5. What 5. What is the ninth root of: 210358 ?

Ans. 3.022239

ARITHMETICAL PROGRESSION.

ANY rank of numbers increasing by a common excess, or decreasing by a common difference, are said to be in Arithmetical Progression ; such are the numbers 1, 2, 3, 4, 5, &c. 7, 5, 3, I; 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 givent, the other two may be readily found.

1. The first term, 2 commonly called the
2. The last term,

extremnes.
3. The number of terms.
4. The common difference.
5. The sum of all the terms.

PROBLEM I. The first term, the last term, and the number of terms being

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.

* Suppose another series of the same kind with the given one be placed ander 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,

EXAMPLES

1. The first term of an arithmetical progression is 2, the last term 53, and the number of terms 18; required the sum of the series.

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2. The first term is 1, the last term 21, and the number of terms 11; required the sum of the series.

Ans. 121.

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

Ans. 300,

4. If

consequently, any one of those sums, multiplied by the number of terms, must give the whole sum of the two series, and half that, sum will evidently be the sum of the given series : thus,

Let 1, 2, 3, 4, 5, 6, 7, be the given series ;

and 7, 6, 5, 4, 3, 2, 1, the same inverted ; then 8+8+8+8+8+8+8=8X7=56 and 1+3+4+5

+6+7=5 = 28. Q. E. D.

4. If 100 stones be placed in a right line, exactly a yard asunder, and the first a yard from a baskot, what length of ground will that man go, who gathers them up singly, returning with them one by one to the basket ?

Ans. 5 miles and 1300 yards.

PROBLEM II.

The first term, the last term, and the number of terms being

give!?,. to find the common difference.

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

EXAMPLES.

1. The extremes are 2 and 53, and the number of terms is 18 ; required the common difference.

18

53

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Or, 53-251

3 the answer. 18-I 17

2. If

* The difference of the first-and last terms evidently shews 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 increase, divided by the number of additions, must give the difference at every one separately ; whence the rule is manifest.

2. If the extremes be 3 and 19, and the number of terms 9, it is required to find the common difference, and the sum of the whole series.

Ans. The difference is 2, and the sum is 99. 3. A man is to travel from London-to a certain place in 12 days, and to go but 3 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 London.

Ans. Daily increase 5, distance 366 miles.

PROBLEM III.

Given the first lerni, the last term, and the common difference,

to find the number of terms.

RULE.*

Pivide the difference of the extremes by the common difference, and the quotient, increased by 1, is the number of terms required.

EXAMPLES.

By the last problem, 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 quotient, augmented by 1, must be the answer to the question.

In any arithmetical progression, the sum of any two of its terms is equal to the sum of any other two terms, taken at an equal distance on contrary sides of the former ; or the double of any one term is equal to the sum of any two terms, taken at an equal distance from it on each side.

The sum of any number of terms (n) of the arithmetical series of odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (12) of that number.

That

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