TO EXTRACT ANY ROOT WHATEVER.*

Let N be the given power or number, n the index of the power, A the assumed power, r its root, R the required root of N.

Then, as the sum of n + 1 times A and n — 1 times N, is to the sum of n + 1 times N and n - 1 times A, so is the assumed root r, to the required root K. Or, as half the said sum of n + 1 times A and n 1 times N, is to the difference between the given and assumed powers, so is the assumed root r, to the difference between the true and assumed roots; which difference, added or subtracted, as the case requires, gives the true root nearly.

That is, n + 1. A + n — 1. N:n+1.N + n - 1. A::r: R.
Or, n + 1. A + n - 1. N:A∽N::r: Ror.

And the operation may be repeated as often as we please, by using always the last found root for the assumed root, and its nth power for the assumed power A.

EXAMPLE,

To extract the 5th root of 21035-8.

Here it appears that the 5th root is between 7.3 and 74. Taking 7-3, its 5th power is 20730-71593. Hence then we have,

N = 21035-8; r = 7·3; n = 5; . n + 1 = 3;.n

A20730-716

N-A305-084

A=20730-716 N = 21035-8

1=2

OF RATIOS, PROPORTIONS, AND PROGRESSIONS.

NUMBERS are compared to each other in two different ways: the one comparison considers the difference of the two numbers, and is named Arithmetical Relation; and the difference sometimes the Arithmetical Ratio: the other considers their quotient, and is called Geometrical Relation, and the quotient the Geometrical Ratio. So, of these two numbers 6 and 3, the difference, or arithmetical ratio, is 6 3 or 3; but the geometrical ratio is for 2.

There must be two numbers to form a comparison: the number which is compared, being placed first, is called the Antecedent; and that to which it is compared, the Consequent. So, in the two numbers above, 6 is the antecedent, and 3 the consequent.

If two or more couplets of numbers have equal ratios, or equal differences, the equality is named Proportion, and the terms of the ratios Proportionals. So, the two couplets, 4,2 and 8,6, are arithmetical proportionals, because 4 =8-6=2; and the two couplets 4,2 and 6,3, are geometrical proportionals, because == 2, the same ratio.

2

To denote numbers as being geometrically proportional, a colon is set between the terms of each couplet, to denote their ratio; and a double colon, or else a mark of equality, between the couplets or ratios. So, the four proportionals, 4, 2, 6, 3, are set thus, 4:2::6:3, which means, that 4 is to 2 as 6 is to 3; or thus, 4:26:3; or thus, * = f, both which mean, that the ratio of 4 to 2, is equal to the ratio of 6 to 3.

Proportion is distinguished into Continued and Discontinued. When the difference or ratio of the consequent of one couplet and the antecedent of the next couplet, is not the same as the common difference or ratio of the couplets, the proportion is discontinued. So, 4, 2, 8, 6, are in discontinued arithmetical proportion, because 4 - 2 = 8 - 6 = 2, whereas, 2-8--6; and 4, 2, 6, 8, are in discontinued geometrical proportion, because == 2, but = 1, which is not the same.

But when the difference or ratio of every two succeeding terms is the same quantity, the proportion is said to be continued, and the numbers themselves a series of continued proportionals, or a progression. So, 2, 4, 6, 8, form an arithmetical progression, because 4 - 26-4= 8 - 6 = 2, all the same common difference; and 2, 4, 8, 16, a geometrical progression, because = = = 2, all the same ratio.

When the following terms of a Progression exceed each other, it is called an Ascending Progression or Series; but if the terms decrease, it is a Descending one.

So, 0, 1, 2, 3, but 9, 7, 5, 3, Also, 1, 2, 4, 8,

and 16, 8, 4, 2,

4, &c., is an ascending arithmetical progression,
1, &c., is a descending arithmetical progression :
16, &c., is an ascending geometrical progression,
1, &c., is a descending geometrical progression.

ARITHMETICAL PROPORTION AND PROGRESSION.

THE first and last terms of a Progression, are called the Extremes; and the other terms, lying between them, the Means.

The most useful part of arithmetical proportions, is contained in the following

'THEOREM 1. If four quantities be in arithmetical proportion, the sum of the two extremes will be equal to the sum of the two means.

Thus, of the four 2, 4, 6, 8, here 2 + 8 = 4 + 6 = 10.

THEOREM 2. In any continued arithmetical progression, the sum of the two extremes, is equal to the sum of any two means that are equally distant from them, or equal to double the middle term when there is an uneven number of terms.

Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 + 3 = 6.

And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 + 12 = 6 + 10 = 8 + 8 = 16.

THEOREM 3. - The difference between the extreme terms of an arithmetical progression, is equal to the common difference of the series multiplied by one less than the number of the terms.

So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the common difference is 2, and one less than the number of terms 9; then the difference of the extremes is 20 2 = 18, and 2 × 9 = 18 also.

Consequently, the greatest term is equal to the least term added to the product of the common difference multiplied by 1 less than the number of terms.

THEOREM 4. The sum of all the terms, of any arithmetical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided by 2; or the sum of the two extremes multiplied by the number of the terms gives double the sum of all the terms in the series.

This is made evident by setting the terms of the series in an inverted order under the same series in a direct order, and adding the corresponding terms together in that order.

in the series

9, 11, 13, 15;
7,
5, 3, 1;

Thus,
1, 3, 5,
7,
ditto inverted, 15, 13, 11, 9,
the sums are 16 +16 +16 +16 +16 +16 +16 +16,

which must be double the sum of the single series, and is equal to the sum of the extremes repeated so often as are the number of the terms.

From these theorems may readily be found any one of these five parts; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given; as in the following Problems :

PROB. I.

Given the extremes, and the number of terms; to find the sum of all the terms. RULE. ADD the extremes together, multiply the sum by the number of terms and divide by 2.

EXAMPLES.

1. The extremes being 3 and 19, and the number of terms 9; required the sum of the terms ?

2. It is required to find the number of all the strokes a clock strikes in one whole revolution of the index, or in 12 hours ? Ans. 78.

3. How many strokes do the clocks of Venice strike in the compass of the

day, which go right on from 1 to 24 o'clock ?

Ans. 300.

4. What debt can be discharged in a year, by weekly payments in arithmetical progression, the first payment being 1s., and the last or 52d payment 5l. 38.? Ans. 1357. 4s.

PROB. II.

Given the extremes, and the number of terms; to find the common difference. RULE-Subtract the less extreme from the greater, and divide the remainder by 1 less than the number of terms, for the common difference.

EXAMPLES.

1. The extremes being 3 and 19, and the number of terins 9; required the common difference?

2. If the extremes be 10 and 70, and the number of terms 21; what is the common difference, and the sum of the series ?

Ans. the com. diff. is 3, and the sum is 840. 3. A certain debt can be discharged in one year, by weekly payments in arithmetical progression, the first payment being 1s., and the last 51. 3s.; what is the common difference of the terms ?

PROB. III.

Ans. 2.

Given one of the extremes, the common difference, and the number of terms; to find the other extreme, and the sum of the series.

RULE-Multiply the common difference by 1 less than the number of terms, and the product will be the difference of the extremes : therefore add the product to the less extreme, to give the greater; or subtract it from the greater to give

the less.

EXAMPLES.

1. Given the least term 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series ?

2

8

16

3

19 the greatest term.

2. If the greatest term be 70, the common difference 3, and the number of terms 21; what is the least term and the sum of the series ?

Ans. the least term is 10, and the sum is 840. 3. A debt can be discharged in a year, by paying ls. the first week, 3s. the second, and so on, always 2s. more every week; what is the debt, and what will the last payment be? Ans. the last payment will be 57. 3s., and the debt is 1354 4s

PROB. IV.

To find an arithmetical mean proportional between two given terms.

RULE. Add the two given extremes or terms together, and take half their sum for the arithmetical mean required. Or, subtract the less extreme from the greater, and half the remainder will be the common difference; which being added to the less extreme, or subtracted from the greater, will give the mean required.

EXAMPLE.

To find an arithmetical mean between the two numbers 4 and 14.

To find two arithmetical means between two given extremes.

RULE.-Subtract the less extreme from the greater, and divide the difference

by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, gives the means.

To find any number of arithmetical means between two given terms or extremes. RULE.-Subtract the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required.

Note. More of Arithmetical Progression is given in the Algebra.