TO EXTRACT THE NINTH ROOT.

RULE.-Extract the cube root of the given number, and you have the third power, whose cube root is the root sought.

What is the ninth root of 5159780352?

Ans. 12.

XXVII. Progression.

When unequal quantities occur, they may be considered in two different points of view: we may inquire, how much greater one quantity is than the other; or we may ask, how many times one quantity is greater than the other. The answer to the first question is found by subtraction, and the answer to the second by division. Although both results are ratios, yet mathematicians have distinguished them into the arbitrary divisions of ARITHMETICAL RATIO, and GEOMETRICAL RATIO. Hence, PROGRESSION is of two kindsARITHMETICAL and GEOMETRICAL.

ARITHMETICAL PROGRESSION.

When a series of numbers increases or decreases by the same quantity, it is called an arithmetical progression. When a series of numbers increases by the same quantity, it is called an ascending series. Thus, 1, 4, 7, 10, 13, 16, 19, 22, &c. are an ascending series of numbers. When a series of numbers decreases by the same number or quantity, it is called a descending series; thus, 22, 19, 16, 13, 10, 7,4, 1. The number which expresses how much greater, or how much less, one term of the series is, than the next term in the same series, is called the difference. Numbers may be employed to denote the number of terms in any series. These numbers are called indices.

Thus, Indices,

1, 2, 3, 4, 5, 6, 7. Arith. Progression, 3, 7, 11, 15, 19, 23, 27, &c.

From these remarks, the propriety of the following rule will be seen.

THE FIRST TERM, THE COMMON DIFFERENCE, AND THE NUMBER OF TERMS BEING GIVEN, TO FIND THE LAST TERM.

RULE.-Multiply the common difference by the number of terms less 1, and to the product add the first term.

1. If the common difference be 4, the first term 3, and the number of terms 5, what is the last term?

We see that this sum may be performed at once by casting the eye upon the indices and series given above. In 5 terms, the common difference is taken 4 times, and the first term once. Therefore, 4X4+3-19, the last term.

2. If the first term be 3, the common difference 4, and the number of terms 7, what is the last term?

In 7 terms, the common difference is taken 6 times, and the first term once. Therefore, 6×4+3=27, last term.

3. If a man has $1 on simple interest, to what sum will it amount in 20 years?

The first term is $1, the common difference $,06, and the number of terms 21. Ans. $2,2.

The number of years is one less than the number of Cerms.*

THE FIRST term, the LAST TERM, AND NUMBER OF TERMS GIVEN, TO FIND THE DIFFERENCE.

RULE.-Subtract the first term from the last, divide the remainder by the number of terms less 1, and the quotient will be the difference.

* It will readily be seen that the years are one less than the number of terms, from the following:

Years,

1 2 3
4

4

5 6 5 6 7

7
8

Arith. Prog. $1, 1,06, 1,12, 1,18, 1,24, 1,30, 1,36, 1,42, &c.

One dollar is a term in the series, but the first year ranges over the second term, $1,06, the third year over the fourth term, and the seventh year over the eighth term.

1. If the first term be 3, the last term 27, and the number of terms 7, what is the difference?

By inspecting the indices and series before given, it will be found that, subtracting 3, the first term from 27 gives the remainder 24, which are composed of the difference taken 6 times. Therefore, 24÷÷6=Ans. 4, difference.

2. If the first term is 4, the last term 39, and the number of terms 8, what is the difference? Ans. 5.

3. A man has 5 sons and 5 daughters, whose several ages differ alike; the youngest of the children was 3 years of age, and the oldest 48. What was the common difference of their ages? 48-3-10-1=5, Ans. 4. A certain school consists of 19 scholars; the youngest is 3 years of age, and the oldest 39. What is the common difference of their ages? Ans. 2 years.

THE FIRST AND LAST TERMS, AND THE COMMON DIFFERENCE GIVEN, TO FIND THE NUMBER OF TERMS.

RULE.-Divide the difference between the first and last terms by the common difference, and to the quotient add 1.

1. If the first term is 3, the last term 27, and the common difference 4, what is the number of terms?

By inspecting the series before given, we shall find that after 3 are subtracted from 27, the remainder 24 is composed of the common difference taken 6 times. Therefore, 27-3÷4+1=7, number of terms. Ans. 7.

As 3, the first term, are subtracted from 27, and the remainder, 24, divided by 4, gives 6, the number of times. which the common difference is taken, 1 must be added to 6, to bring into the computation the first term, 3.

2. If the first term is 3, the common difference 4, and the last term 35, what is the number of terms? Ans. 9.

3. If the first term is 4, the common difference 2, and the last term 22, what is the number of terms? Ans. 10.

4. If the first term is 4, the common difference 2, and the last term 14, what is the number of terms? Ans. 6.

5. If the first term is 5, the common difference 4, and the last term 45, what is the number of terms? Ans. 11.

6. If the first term is 33, the last term 73, and the common difference 13, what is the number of terms? Ans. 4. 7. If the first term is 2, the last term 6, and the common difference 14, what is the number of terms? Ans. 4.

THE FIRST AND LAST TERM, AND THE NUMBER OF TERMS GIVEN, TO FIND THE SUM OF ALL THE TERMS.

RULE.-Multiply the sum of the first and last term by the number of terms, and half the product will be the sum of all the terms. Or, which is equally correct, multiply the sum of the first and last term by half the number of terms.

1. If the first term is 2, the last term 29, and the number of terms 10, what is the sum of all the terms? Ans. 155.

ILLUSTRATION.

2+ 5+ 8+11+14+17+20+23+26+29

29+26+23+20+17+14+11+ 8+ 5+ 2

31+31+31+31+31+31+31+31+31+31=

310-2=155.

Or, which amounts to the same, 31×10=310÷÷2=155. 2. If the first term is 3, the last term 45, and the number of terms 22, what is the sum of all the terms? Ans. 528.

3. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term? Ans. 301.

4. There are, in a certain triangular field, 41 rows of corn; the first row, in one corner, is a single hill; the second contains 3 hills, and so on, with a common difference of two; what is the number of hills in the last row?

Ans. 81 hills.

5. If 100 stones be placed in a straight line, at the distance of a yard from each other, how far must a person travel to bring them one by one to a box placed at the distance of a yard from the first stone? Ans. 5 mi. and 1300 yds.

GEOMETRICAL PROGRESSION.

Numbers, increasing or decreasing by a certain number of times, form a geometrical progression. Or, quantitie are in geometrical progression when they increase by the same multiplier, or decrease by the same divisor.

The multiplier or divisor is called the ratio. When numbers increase by the same multiplier, they form an ascending series; when they decrease by the same divisor, they form a descending series.

THE RATIO, NUMBER OF TERMS, AND one extreme given, TO FIND THE OTHER EXTREME.

RULE.-When the series is ascending, multiply the given extreme by such power of the ratio, whose index is equal to the number of terms less 1, and the product will be the other

extreme.

When the series is descending, divide the given extreme by such power of the ratio, whose index is equal to the number of terms less 1, and the quotient will be the other extreme.

The reason of the rule is evident from the manner in which a geometrical progression is formed. Let 2, 4, 8, 16, 32, &c. be the series whose ratio is 2. The second term is formed by multiplying the first by the ratio; the third term by multiplying the second by the ratio, and so on. The series may therefore be written thus: 2, 2×2, 2×2×2, 2×2×2×2, 2×2×2×2×2, &c. Any term after the first is evidently that power of the ratio whose index is one less than the number of the term multiplied by the first term. Thus, the third term is 2x22; the fourth term is 2×23; and the eighth term would be 2×27, and so on. In an ascending series, therefore, multiply the first term by that power of the ratio, whose index is one less than the number of the term sought, and the product is the term sought. In a descending series, as, 243, 81, 27, 9, 3, 1, whose ratio is 3, and which is also 1X35, 1X3a, 1×33, 1X35. 1x32, 1×31, 1, the last term, 1=!

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