Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

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.

EXAMPLES.

1. What is the cube root of 53157376?

53157376(376

27=33

3X327)261 dividend.

50653=373

3*X34107)25043 second dividend,
53157376

2. What is the biquadrate root of 19987173376 ?

Ans. 376

3. Extract the sursolid, or fifth root, of 307682821106

715625.

Ans. 3145.

4. Extract

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

Ans. 27534.

[blocks in formation]

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 I and the given number multiplied by the index less 1, is to the sum of the given number multiplied by the index more 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

Here it appears that the fifth root is between 7'3 and 1:4. 73 being taken, its fifth power is 20730'71593*

Hence then

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

2. What is the third root of 2?

3. What is the sixth root of 21035.8?

Ans, 1259921.

Ans. 5*254037.

4. What is the seventh root of 210358?

Ans. 4*145392.

5. What

5. What is the ninth root of 21035'8 ?

Ans. 3022239.

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 given, the other two may be readily found."

[blocks in formation]

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 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,

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.

[blocks in formation]

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=50=28. Q. E. D.

X

« ΠροηγούμενηΣυνέχεια »