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

This rule will be sufficiently obvious from the work in the following example.

Extract the cube root of a + 6a3—40a3 +96α-64. a® + 6a5—40a3 +96a—64(a2 +2u-4

a2+2a×3=3a* +12a3 +12a 2)—12aa—48a3+96a-64(-4

a+6a5—40a3+96a—64a2+2a-4

3

When the index of the power, whose root is to be extracted, is a composite number, the following rule will be serviceable :

Take any two or more indices, whose product is the given index, and extract out of the given number a root answering to one of these indices; and then out of this root extract a root, answering to another of the indices, and so on to the last.

Thus, the fourth root = square root of the square root.

The sixth root = square root of the cube root, &c.

The proof of all roots is by involution.

The following theorems may sometimes be found useful in ex

tracting the root of a vulgar fraction;====

a

2. Find the first figure of the root by trial, and subtract its power from the given number.

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 inferior 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, till the whole be finished.

3. Extract the sursolid, or fifth root, of 30768282110671 5625.

Ans. 3145.

4. Extract the square cubed, or sixth root, of 43572838 1009267809889764416.

Ans. 27534.

5. Find the seventh root of 3448771746730751318249 2153794673.

Ans. 32017.

6. Find the eighth root of 11210162813204762362464

97942460481.

Ans. 13527.

TO EXTRACT ANY ROOT WHATEVER BY

APPROXIMATION.

RULE.

1. Assume the root nearly, and raise it to the same power with the given number, which call the assumed power.

2. Then, as the sum of the assumed power multiplied by the index more 1 and the given number multiplied by the index less 1, is to the sum of the given number multiplied by the index more 1 and the assumed power multiplied by the index less 1, so is the assumed root to the required root.

Or, as half the first sum is to the difference between the given and assumed powers, so is the assumed root to the difference between the true and assumed roots; which difference, added or subtracted, gives the true root nearly.

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

EXAMPLES.

1. Required the fifth root of 21035'8.

Here it appears, that the fifth root is between 7°3 and 74. 73 being taken, its fifth power is 20730'71593. Hence then

104263*7)2227*1132('0213604 = difference.

ARITHMETICAL PROGRESSION.

ANY rank of numbers, increasing by a common excess or decreasing by a common difference, is said to be in Arith

metical Progression; such are the numbers 1, 2, 3, 4, 5, &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 be readily found.

1. The first term,

2. The last team,

}

commonly called the

3. The number of terms.

4. The common difference.

5. The sum of all the terms.

extremes.

PROBLEM 1.

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.

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

* Suppose another series of the same kind with the given one 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 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=28. Q. E. D.

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