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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 is contained in the dividend, and the quotient will be another figure of the root. An allowance of two or three units is generally made, and for the higher powers a still greater allowance is necessary.

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, as before; and in like manner proceed till the whole is finished.

1. What is the cube root of 20346417 ?

OPERATION. 20346417(273

= lst subtrahend. 22 X 3= 12) 123 = lst dividend. 273 =

= 2d subtrahend. 272 X 3 = 2187) 6634 = 2d dividend 2733 =

20346417 = 3d subtrahend.

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2. What is the fourth root of 34828517376 ?

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OPERATION.
34828517376(432 Ans.
256

= Ist subtrahend. 43 x 4 = . 256) 922

= lst dividend. 434 =

3418801

-= 2d subtrahend. 433' X 4 = 318028) 640507 = 2d dividend. 4324 =

34828517376 = 3d subtrahend. 3. What is the 5th root of 281950621875 ? Ans. 195. 4. Required the sixth root of 1178420166015625.

Ans. 325. 5. Required the seventh root of 1283918464548864.

Ans. 144. 6. Required the eighth root of 218340105584896.

Ans. 62.

Section LXIII.
ARITHMETICAL PROGRESSION.

When a series of quantities or numbers increases or de. creases by a constant difference, it is called arithmetical progression, or progression by difference. The constant difference is called the common difference or the ratio of the progression. Thus, let there be the two following series:

1, 5, 9, 13, 17, 21, 25, 29, 33,

25, 22, 19, 16, 13, 10, 7, 4, 1. The first is called an ascending series or progression. The second is called a descending series or progression.

The numbers which form the series are called the terms of the progression.

The first and last terms of the progression are called the extremes, and the other terms, the means.

Any three of the five following things being given, the other two may be found :

1st, the first term,
2d, the last term,
3d, the number of terms,
4th, the common difference,
5th, the sum of the terms.

PROBLEM I. The first term, last term, and number of terms being given, to find the common difference.

To illustrate this problem, we will examine the following series, –

3, 5, 7, 9, 11, 13, 15, 17, 19. It will be perceived that in this series 3 and 19 are the extremes, 2 the common difference, 9 the number of terms, and 99 the sum of the series.

It is evident, that the number of common differences in any number of terms will be one less than the number of terms. Hence, if there be 9 terms, the number of common differences will be 8, and the sum of these common differences will be equal to the difference of the extremes ; therefore if the difference of the extremes, 19 — 3 = 16, be divided by the number of common differences, the quotient will be the common difference. Thus 16 -8= 2 is the common difference.

Rule. Divide the difference of the extremes by the number of terms less one, and the quotient is the common difference.

1. The extremes are 3 and 45, and the number of terms is 22. What is the common difference?

OPERATION. 45- 3

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2. A man is to travel from Albany to a certain place in 11 days, and to go but 5 miles the first day, increasing the distance equally each day, so that the last day's journey may be 45 miles. Required the daily increase. Ans. 4 miles.

3. A man had 10 sons, whose several ages differed alike; the youngest was 3 years, and the oldest 48. What was the common difference of their ages ?

Ans. 5 years. 4. A certain school consists of 19 scholars; the youngest is 3 years old, and the oldest 39. What is the common difference of their ages ?

Ans. 2 years.

PROBLEM II. The first term, last term, and number of terms being given, to find the sum of all the terms.

Illustration. Let 3, 5, 7, 9, 11, 13, 15, 17, 19, be the series, and 19, 17, 15, 13, 11, 9, 7, 5, 3, the same series inverted.

22, 22, 22, 22, 22, 22, 22, 22, 22, sum of both series. From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series.

It is evident that if, in the above series, 22 be multiplied by 9, the number of terms, the product will be the sum of both series, 22 X 9 = 198, and therefore the sum of either series will be 198 • 2= 99. But 22 is also the sum of the extremes in either series, 3 + 19 = 22. Therefore, if the sum of the extremes be multiplied by the number of terms, the prod. uct will be double the sum of the series.

Rule. -.. Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series.

5. The extremes of an arithmetical series are 3 and 45, and the number of terms 22. Required the sum of the series.

OPERATION. 45+3 X 22

-=528 Answer.

2 6. A man going a journey travelled the first day 7 miles, the last day 51 miles, and he continued his journey 12 days. How far did he travel ?

Ans. 348 miles. 7. In a certain school there are 19 scholars ; the youngest is 3 years old, and the oldest 39. What is the sum of their ages ?

Ans. 399 years. 8. Suppose a number of stones were laid a rod distant from each other, for thirty miles, and the first stone a rod from a basket. What length of ground will that man travel over who gathers them up singly, returning with them one by one to the basket ?

Ans. 288090 miles 2 rods.

PROBLEM III. The extremes and the common difference being given, to find the number of terms.

Illustration. — Let the extremes be 3 and 19, and the common difference 2. The difference of the extremes will be 19 - 3

= 16; and it is evident, that, if the difference of the extremes be divided by the common difference, the quotient is the number of common differences ; thus 16 ; 2=8. We have de. monstrated in Problem 1. that the number of terms is one more than the number of differences; therefore 8+1=9, the number of terms.

RULE. – Divide the difference of the extremes by the common difference, and the quotient increased by one will be the number of terms required.

9. If the extremes are 3 and 45, and the common difference 2, what is the number of terms?

OPERATION.
45 -3 11-02 Answer.

2

10. In a certain school the ages of all the scholars differ alike ; the oldest is 39 years, the youngest is 3 years, and the difference between the ages of each is 2 years. Required the number of scholars. .

Ans. 19. 11. A man going a journey travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles. How many days did he travel ?

Ans. 12.

PROBLEM IV. The extremes and common difference being given, to find the sum of the series.

Illustration. — Let the extremes be 3 and 19, and the common difference 2. The difference of the extremes will be 19

-3= 16; and it has been shown in the last problem, that, if the difference of the extremes be divided by the common difference, the quotient will be the number of terms less one ; therefore the number of terms less one will be 16 ; 2= 8, and the number of terms 8+1=9. It was demonstrated in Problem II. that if the number of terms was multiplied by the sum of the extremes, and the product divided by 2, the quotient would be the sum of the series.

Rule. - Divide the difference of the extremes by the common difference, and add 1 to the quotient; multiply this quotient by the sum of the extremes, and half the product is the sum of the series.

12. If the extremes are 3 and 45, and the common differ. ence 2, what is the sum of the series ?

Ans. 528. 13. A owes B a certain sum, to be discharged in a year, by paying 6 cents the first week, 18 cents the second week, and thus to increase every week by 12 cents, till the last payment should be $ 6.18. What is the debt ? Ans. $162.24.

PROBLEM V. The extremes and sum of the series being given, to find the common difference.

Illustration. — Let the extremes be 3 and 19, and the sum of the series 99, to find the common difference. We have before shown, that, if the extremes be multiplied by the number of terms, the product would be twice the sum of the series ; there. fore, if twice the sum of the series be divided by the extremes, the quotient will be the number of terms. Thus, 99 X 2= 198; 3+ 19 = 22; 198 ; 22 = 9 is the number of terms. And we have before shown, that, if the difference of the extremes be divided by the number of terms less one, the quotient will be the common difference; therefore, 19—3= 16; 9-1=8; 16 -8= 2 is the common difference.

Rule. – Divide twice the sum of the series by the sum of the extremes, and from the quotient subtract 1; and with this remainder divide the difference of the extremes, and the quotient is the common difference.

14. The extremes are 3 and 45, and the sum of the series 528. What is the common difference?

Ans. 2.

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