been given for extracting the cube root, depends. In this general application of the principle, the given number is to be divided into periods, each consisting of as many figures as correspond to the order of the root, and the same number of columns is to be employed, the last headed by the given number, and the others by ciphers. The work then proceeds exactly as in the extraction of the cube root; and if there be a remainder, a like contraction is admissible with equal advantage.

Exam. 1. Required the fifth root of 95.

In this example, as the root is of the fifth order, we have five columns, the first four commencing with zeros, and the fifth with the given number, the fraction being reduced to a decimal carried out to five places, the number constituting the period in extracting the fifth root, and the last being made 2, because of the 8 which would follow it. Then, the first figure of the root is 2, since the fifth power of 2 is 32, while that of 3 is too great, being 243. Now, after a process exactly analogous to that employed in extracting the cube root, we find, that, by adding one cipher in the first column, two in the second, &c., the numbers in the several columns which are to be employed in working for the second figure are 100, 4000, 80000, 800000, and 6318182. The next figure is found to be 4; and, after the usual process, the numbers in the several columns, preparatory to the work for the next figure, turn out to be 120,

5760, 138240, 1658880, and 1555558. Then, in commencing the contraction, we cut one figure from the fourth column, two from the third, three from the second, and four from the first (which last is thus exhausted). The rest of the work presents no difficulty.

This method of extracting roots is perfect in principle, and in most of the useful cases of evolution, it gives, perhaps, as easy solutions as the nature of the case will admit. In some instances, however, especially when the roots are of a high order, the process will be easier by means of the following approximate rule, first given by Dr. Hutton, in the tenth of whose Tracts its investigation may be found :—

RULE II. To extract any root whatever: (1.) Call the index of the given power n; and find by trial a number nearly equal to the required root, and call it the assumed root. (2.) Raise the assumed root to the power whose index is n. (3.) Then, as n+1 times this power, added to n-1 times the given number, is to n-1 times the same power added to n+1 times the given number, so is the assumed root to the true root nearly. (4.) The number thus found may be employed as a new assumed root, and the operation repeated to find a result still nearer the true root.

Exam. 2. Required the 365th root of 1'06.

Here we may take 1 for the assumed root, the 365th power of which is 1; and n being 365, we have n+1=366, and n-1=364. The work will then proceed in the following manner, and the answer will be found to be 1·0001596.

In extracting the fourth root, we may either use one of the preceding rules (Rule I. will be preferable), or we may extract the second root of the given number, and the second root of the result. In extracting the sixth root, also, we may either employ one of those rules, or we may extract the third root of the given number, and the second root of the result: and in this way we may proceed in every case in which the index of the root to be extracted is a composite number. When the index is a prime number, however, the root must be found by one of the general rules.

Exam. 3. Find the value of 113.

This expression means the third root of the second power of 11, and, therefore, by extracting the cube root of 121, we find for the required result 4.946088. The answer might also be obtained by the second rule, by taking 5 for the assumed root, and, the reciprocal of, for n. The former method is preferable.

N

Exam. 4. From the data in Example 10, page 235, let it be required to find the annual rate, per cent., of the increase of the population of Glasgow, between the years 1801 and 1811, supposing the increase to be always proportional to the population.

Here, by dividing the greater population by the less, we get 1-301909, of which, because the interval is ten years, the tenth root is to be taken. This is effected most easily by first extracting the square root, which is found to be 1-141017; the fifth root of which (found very easily by Rule 1.) is 102674. Then, multiplying this by 100, and rejecting 100 from the product, we get 2.674, the annual rate per cent. required. The reason of the process will be understood by those who have studied compound interest.

Exercises. Find the roots of the following numbers signified by their several indices:

A SERIES is a succession of quantities, or terms, that depend on one another according to a certain law.

In every series, the first and last terms are called the EXTREMES, and the rest the MEANS.

Writers on arithmetic usually treat of only two kinds of series, equidifferent series and continual proportionals. These are of more frequent and general use than other kinds of series, and, on this account, claim more particular attention. Quantities in equidifferent series are also said to be in arithmetical progression; and continual proportionals are said to be in geometrical progression.*

*These names for equidifferent quantities and continual proportionals are very improper. Series of both kinds belong equally to arithmetic and geometry. The appellations, arithmetical progression and geometrical progression, should, therefore, be entirely disused, as tending to impress false ideas on the mind respecting the nature of the quantities. The term proportion is applied, if possible, still more improperly to equidifferent quantities, as this term is always expressive not of equality of differences, but of equality of ratios. The latest and best Continental writers have accordingly rejected these terms, and substituted more appropriate ones, calling them by the names above given, or others of similar import, such as progressions by differences, and progressions by quotients. With regard to the name, continual proportionals, here applied to the second kind of quantities, it may be observed, that, besides its being perfectly expressive of the nature of such quantities, it has long been thus applied in works on geometry; and it is equally applicable in arithmetic.

EQUIDIFFERENT SERIES, OR ARITHMETICAL PRO-
GRESSION.

When three or more numbers or quantities are such, that each of them, after thè first, exceeds the one before it, or that each is less than the one before it, by a constant number or quantity, called the COMMON DIFFERENCE, they are said to be EQUIDIFFERENT, or to be in ARITH

METICAL PROGRESSION.

When each term, after the first, is greater than the one before it, the series is said to be an ASCENDING one; otherwise, it is a DESCENDING one.

Thus, 5, 7, 9, 11, 13, is an ascending equidifferent series, in which each term is derived from the one before it, by the addition of the common difference 2; and 20, 17, 14, 11, 8, is a descending one, in which each term is less than the one before it, by the common difference 3.

The following are the most useful rules for the management of quantities of this kind :—

RULE I. The first term and the common difference being given, to find any other assigned term: (1.) Multiply the common difference by the number which is equal to the number of terms preceding the required term. (2.) Then, if the series be an ascending one, add the product to the first term; otherwise, subtract it.

Exam. 1. Required the thirty-fifth term of the increasing equidifferent series, whose first term is 7, and common difference 3.

=

Here, 34 terms precede the required one; wherefore, 34 × 3 +7 = 109 is the term required.

The reason of this operation will be manifest from the consideration, that were the series to be continued to the thirty-fifth term, the first term must be increased by thirty-four additions of the common difference.

Exer. 1. Required the fifty-fourth term of the decreasing equidifferent series whose first term is 100, and common difference 11. Answ. 333

2. Given the first term of an increasing series = 36, and the common difference = 3; to find the hundredth term. Answ. 392. 3. If the first term of a decreasing series be 329, and the common difference }; what is the ninety-ninth term? Answ. 2431.

RULE II. The extremes and the number of terms being given, to find the sum of the series: Multiply the sum of the extremes by the number of terms, and take half the product.

Exam. 2. The first term of an equidifferent series is 1, its last term 312, and the number of terms 193. What is its sum?

Here, 1+312 313, and 313 × 193 = 60409; the half of which is 30204, the sum of the series.

The reason of this rule will be understood from the following property of equidifferent quantities:

In any equidifferent series, the sum of the extremes is equal to the sum of any two terms that are equally distant from them, or to double the middle term, if the number of terms be odd.

Thus, in the series, 5, 8, 11, 14, 17, 20, 23, the sum of 5 and 23 is equal to the sum of 8 and 20, or of 11 and 17, and is double of the middle term 14. The reason of this is plain, since 8 and 11 respectively exceed the less extreme by the same quantities by which 20 and 17 are respectively less than the other extreme.

Hence, in this latter series, it is evident, that though each term were made 14, half the sum of the extremes, still the sum of the whole would be the same; and consequently the sum of the series will be obtained by multiplying half the sum of the extremes by the number of terms, or, which is the same, by multiplying the sum of the extremes by the number of terms, and taking half the product.

Exer. 4. Given the greater extreme = 1000, the common difference=21, and the number of terms 367; required the sum of the series. Answ. 221484.

In working this exercise, the less extreme will be found by Rule I., and the sum by Rule II.

=

5. Given the greater extreme-1, the common difference and the number of terms = 51; required the sum of the series. Answ. 38.

RULE III. The extremes and the common difference being given, to find the number of terms: Divide the difference of the extremes by the common difference, and add a unit to the quotient.

The reasons of this rule and of the next will be obvious from comparing them with Rule I.

Exer. 6. If the greater extreme be 500, the less 70, and the common difference 10; what is the number of terms? Answ. 44.

7. Given the less extreme = 3, the greater = 579, and the common difference = 9, to find the sum of the series. Answ. 18915.

Here, let the number of terms be found by this rule, and the sum of the series by Rule II.

8. With the common difference 12, how many equidifferent means can be inserted between the extremes 8 and 1700? Answ. 140.

RULE IV. The extremes and the number of terms being given, to find the common difference: Take 1 from the