171. A SERIES, in algebra, consists of an infinite number of terms following one another, each of which is derived from one or more of the preceding ones by a fixed law. This law is called the law of the series.

Arithmetical Progression.

172. An ARITHMETICAL PROGRESSION is a series, in which each term is derived from the preceding one by the addition of a constant quantity called the common difference.

If the common difference is positive, each term will be greater than the preceding one, and the progression is said to be in creasing.

If the common difference is negative, each term will be less than the preceding one, and the progression is said to be decreasing.

Thus, . . . 1, 3, 5, 7, . . . &c., is an increasing arithmetical progression, in which the common difference is 2;

and

19, 16, 13, 10, 7, . . . is a decreasing arithmetical progression, in which the common difference is 3.

173. When a certain number of terms of an arithmetical progression are considered, the first of these is called the first term of the progression, the last is called the last term of the progression, and both together are called the extremes. All the terms between the extremes are called arithmetical means. An arithmetical progression is often called a progression by differences.

174. Let d represent the common difference of the arithmetical progression,

which is written by placing a period between each two of the terms.

From the definition of a progression, it follows that,

b=a+d, c=b+da+2d, e = c + a = a + 3d;

and, in general, any term of the series, is equal to the first term plus as many times the common difference as there are preceding terms.

Thus, let be any term, and n the number which marks the place of it. Then, the number of preceding terms will be denoted by n 1, and the expression for this general term, will be

If d is positive, the progression will be increasing; hence, In an increasing arithmetical progression, any term is equal to the first term, plus the product of the common difference by the number of preceding terms.

If we make n = 1, we have la; that is, there will be but one term.

If we make

n = 2, we have la+d;

that is, there will be two terms, and the second term is equal to the first plus the common difference.

EXAMPLES,

1. If a 3 and d = 2, 2. If a 5 and d=4,

what is the 3d term?
what is the 6th term?

Ans. 7.

Ans. 25.

=

3. If a 7 and d 5. what is the 9th term?

Ans. 47.

serves to find any term whatever, without determining those

which precede it.

Thus, to find the 50th term of the progression,

174*. If d is negative, the progression is decreasing, and the formula becomes

Any term of a decreasing arithmetical progression, is equal to the first term plus the product of the common difference by the number of preceding terms.

EXAMPLES.

1. The first term of a decreasing progression is 60, and the common difference 3: what is the 20th term?

l=a — (n − 1) d gives 760 (201) 3 = 60 — 57 = 3.

2. The first term is 90, the common difference Is the 15th term?

4: what Ans. 34.

2.

Ans. 22.

3. The first term is 100, and the common difference what is the 40th term?

175. If we take an arithmetical progression,

having n terms, and the common difference d, and designate the term which has p terms before it, by t, we shall have

If we revert the order of terms of the progression, considering as the first term, we shall have a new progression whose common difference is - d. The term of this new progression which has p terms before it, will evidently be the same as that which has p terms after it in the given progression, and if we represent that term by t', we shall have,

Adding equations (1) and (2), member to member, we find

t + t' = a + 1; hence,

The sum of any two terms, at equal distances from the extremes of an arithmetical progression, is equal to the sum of the extremes.

176. If the sum of the terms of a progression be represented by S, and a new progression be formed, by reversing the order of the terms, we shall have

Adding these equations, member to member, we get

28 = (a+1)+ (b + k) + (c + i) . . . + (i + c) + (k + b) + (1 + a) ;

and, since all the sums, a + 1, b + k, c+i

are equal to each other, and their number equal to n, the number of terms in the progression, we

have

· ( a + 1) n

n; that is,

The sum of the terms of an arithmetical progression is equal to half the sum of the two extremes multiplied by the number of terms.

EXAMPLES.

1. The extremes are 2 and 16, and the number of terms 8: what is the sum of the series?

2. The extremes are 3 and 27, and the number of terms 12: what is the sum of the series?

Ans. 180.

3. The extremes are 4 and 20, and the number of terms 10: what is the sum of the series?

Ans. 120.

4. The extremes are 8 and 80, and the number of terms 10: what is the sum of the series?

Ans. 440.

contain five quantities, a, d, n, 1, and S, and consequently give rise to the following general problem, viz. :

Any three of these five quantities being given, to determine the other two.

This general problem gives rise to the ten following cases :

No. Given. Unknown.j

Values of the Unknown Quantities.

1a, d, nl, Sla + (n-1) d; Sn [2a + (n − 1)d].

(l + a) (l — a + d)

2d

7d, n, la, Sal− (n − 1)d; S = {n [21 − (n − 1) d].

The first term of an increasing arithmetical progression, is equal to any following term, minus the product of the common difference by the number of preceding terms.

178. From the same formula, we also find

that is,