ART. 223.-Given, the first term a, the common difference d, and the number of terms n, to find s, the sum of the series.

If we take an arithmetical series of which the first term is 3, common difference 2, and number of terms 5, it may be written in the following forms:

3, 3+2, 3+4, 3+6, 3+8

11, 11–2, 11-4, 11-6, 11–8.

It is obvious, that the sum of all the terms in either of these lines, will represent the sum of the series; that is,

8=3+( 3+2)+( 3+4)+( 3+6)+(3+8)
S=11+(11-2)+(11-4)+(11—6)+(11—8)

And

Adding, 2s 14+ 14

=14X5=70

Whence, s of 70–35.

Now, let

the last term, then writing the series both in a

direct and inverted order,

And

s=a+(a+d)+(a+2d)+(a+3d)+...
8=1+(l―d)+(l—2d)+(1-3d)+

By adding the corresponding terms, we have

the series.

28=(1+a)+(1+a)+(l+a)+(l+a).

• +(i+a)

=(1+a) taken as many times as there are terms (n) in

FOR FINDING THE SUM OF AN ARITHMETICAL SERIES.

Multiply half the sum of the two extremes, by the number of terms. From the preceding, it appears, that the sum of the extremes is equal to the sum of any other two terms equally distant from the extremes.

REMARK.—Since l—a+(n−1)d, if we substitute this in the place of l This gives

n

in the formula s=(1+a), it becomes s= ( 2a+(n−1)d ), .

the following Rule, for finding the sum of an arithmetical series: To the double of the first term add the product of the number of terms less one, by the common difference, and multiply the sum by half the number of terms.

REVIEW.-223. What is the rule for finding the sum of an arithmetical series? Explain the reason of the rule.

EXAMPLES.

1. Find the sum of an arithmetical series, of which the first term is 3, last term 17, and number of terms 8.

8=

(3+17) 8-80. Ans.

2

2. Find the sum of an arithmetical series, whose first term is 1, last term 12, and number of terms 12. Ans. 78. 3. Find the sum of an arithmetical series, whose first term is 0, common difference 1, and number of terms 20. Ans. 190. 4. Find the sum of an arithmetical series, whose first term is 3, common difference 2, and number of terms 21.

5. Find the sum of an arithmetical series, whose 10, common difference -3, and number of terms 10.

Ans. 483.

first term is

A. -35.

In this case, the sum of the negative terms exceeds that of the positive.

ART. 224.—The equations l=a+(n−1)d and

n

s=(a+1), furnish the means of

solving this general problem: Knowing any three of the five quantities a, d, n, l, s, which enter into an arithmetical series, to determine the other two.

This question furnishes ten problems, the solution of which presents no difficulty; for we have always two equations, to determine the two unknown quantities, and the equations to be solved, are either those of the first or second degree.

1. Let it be required to find a in terms of l, n, and d.

From the first formula, by transposing, we have a⇒l—(n—1)d. That is, the first term of an increasing arithmetical series is equal to the last term diminished by the product of the common difference into the number of terms less one.

From the same formula, by transposing a, and dividing by n−1, 1-a we find d-

n-l'

That is, in any arithmetical series, the common difference is equal to the difference of the extremes, divided by the number of terms less one. Examples, illustrating these principles, will be found in the collection at the close of this subject.

REVIEW.-224. What are the fundamental equations of arithmetical progression, and to what general problem do they give rise? To what is the first term of an increasing arithmetical series equal? To what is the common difference of an arithmetical series equal?

ART. 225.-By means of the preceding principle, we are enabled to solve the following problem.

Two numbers, a and b, being given, to insert a number, m, of arithmetical means between them; that is, so that the numbers inserted, shall form, with the two given numbers, an arithmetical series.

Regarding a and b as the first and last terms of an increasing arithmetical series, if we insert m terms between them, we shall have a series consisting of m+2 terms. But, by the preceding principle, the common difference of this series will be equal to the difference of the extremes divided by the number of terms less b-a b-a one; that is, d= ; therefore, the common difference ̄m+2−1 ̄m+1

=

will be equal to the difference of the two numbers, divided by the number of means plus one.

Let it be required to insert five arithmetical means between 3 and 15.

15-3 Here d= 5+1

=2; hence the series is 3, 5, 7, 9, 11, 13, 15. It is evident, that if we insert the same number of means between the consecutive terms of an arithmetical series, the result will form a new progression. Thus, if we insert 3 terms between the consecutive terms of the progression, 1, 9, 17, &c., the new series will be 1, 3, 5, 7, 9, 11, 13, 15, 17, and so on.

EXAMPLES.

1. Find the sum of the natural series of numbers 1, 2, 3, 4, . earried to 1000 terms. Ans. 500500.

2. Required the last term, and the sum of the series of odd numbers 1, 3, 5, 7, continued to 101 terms.

Ans. 201 and 10201. 3. How many times does a common clock strike, in a week?

Ans. 1092.

4. Find the nth term, and the sum of n terms of the natural series of numbers 1, 2, 3, 4 . . . Ans. n, and n(n+1). 5. Find the nth term, and the sum of n terms, of the series of odd numbers 1, 3, 5, 7. Ans. 2n-1, and n2.

6. The first and last terms of an arithmetical series are 2 and 29, and the common difference is 3; required the number of terms and the sum of the series. Ans. 10 and 155.

7. The first and last terms of a decreasing arithmetical series are 10 and 6, and the number of terms 9; required the common difference, and the sum of the series. Ans. and 72.

8. The first term of a decreasing arithmetical series is 10, the number of terms 10, and the sum of the series 85; required the last term and the common difference. Ans. 7 and

9. Required the series obtained from inserting four arithmetical means between each of the two terms of the series 1, 16, 31, &c. Ans. 1, 4, 7, 10, 13, 16, &c. 10. The sum of an arithmetical progression is 72, the first term is 24, and the common difference is -4; required the number of terms. Ans. 9 or 4.

In finding the value of n in this question, it is required to solve the equation n2—13n=-36, which has two roots, 9 and 4. These give rise to the two following series, in both of which the sum is 72.

First series,

24, 20, 16, 12, 8, 4, 0, —4, −8. Second series, 24, 20, 16, 12.

11. A man bought a farm, paying for the first acre 1 dollar, for the second 2 dollars, for the third 3 dollars, and so on; when he came to settle, he had to pay 12880 dollars; how many acres did the farm contain, and what was the average price per acre?

Ans. 160 acres, at $80 per acre.

12. If a person, A, start from a certain place, traveling a miles the first day, 2a the second, 3a the third, and so on; and at the end of 4 days, B start after him from the same place, traveling uniformly 9a miles a day; when will B overtake A?

Let x= the number of days required; then the distance traveled by A in x days =a+2a+3a, &c., to x terms,

ax(x+1); and

the distance traveled by B in (x—4) days =9a(x—4). Whence ax(x+1)=9a(x-4). From which x 8, or 9.

Hence, B overtakes A at the end of 8 days; and since, on the ninth day, A travels 9a miles, which is B's uniform rate, they will be together at the end of the ninth day. This is an instance of the precision with which the solution of an equation points out the circumstances of a problem.

13. A sets out 3 hours and 20 minutes before B, and travels at the rate of 6 miles an hour; in how many hours will B overtake A, if he travel 5 miles the first hour, 6 the second, 7 the third, and so on? Ans. 8 hours.

14. Two travelers, A and B, set out from the same place, at the same time. A travels at the constant rate of 3 miles an hour, but B's rate of traveling, is 4 miles the first hour, 31⁄2 the second, 3 the third, and so on, in the same series; in how many hours will A overtake B? Ans. 5 hours. REVIEW.-225. How do you insert m arithmetical means between two given numbers ?

GEOMETRICAL PROGRESSION.

ART. 226.—A Geometrical Progression is a series of terms, each of which is derived from the preceding, by multiplying it by a constant quantity, termed the ratio.

Thus, 1, 2, 4, 8, 16, &c., is an increasing geometrical series, whose common ratio is 2.

Also, 54, 18, 6, 2, &c., is a decreasing geometrical series, whose common ratio is }.

Generally, a, ar, ar3, ar3, &c., is a geometrical progression, whose common ratio is r, and which is an increasing or decreasing series, according as r is greater, or less than 1.

It is obvious, that the common ratio in any series, will be ascertained by dividing any term of the series, by that which immediately precedes it.

REMARK. A geometrical progression is termed, by some writers, an quirational series, or a series of continued proportionals, or a progression by quotients.

ART. 227.-To find the last term of the series.

Let a denote the first term, r the common ratio, 7 the nth term, and s the sum of n terms; then, the respective terms of the series will be

That is, the exponent of r in the second term is 1, in the third term 2, in the fourth term 3, and so on; hence, the nth term of the series will be, lar"-1; that is,

Any term of a geometric series is equal to the product of the first term, by the ratio raised to a power, whose exponent is one less than the number of terms.

EXAMPLES.

1. Find the 5th term of the geometric progression, whose first term is 4, and common ratio 3.

3*=3×3×3×3=81, and 81X4=324, the fifth term. 2. Find the 6th term of the progression 2, 8, 32, &c.

Ans. 2048.

3. Given the 1st term 1, and ratio 2, to find the 7th term.

Ans. 64.

4. Given the 1st term 4, and ratio 3, to find the 10th term.

Ans. 78732. Give examples of How may the com

REVIEW.-226. What is a Geometrical Progression? an increasing, and of a decreasing geometrical series. mon ratio in any geometrical series be found? 227. How is any term of a geometrical series found? Explain the principle of this rule.