In a series of numbers, as 1, 3, 5, 7, 9, etc., proceeding in regular order, increasing by a common difference, the series is called an arithmetical progression; the quantity by which one number is increased beyond the preceding one is termed the difference. If d represent the difference and a the first term, then the progression may be stated thus -

Terms -

1,

2,

3,

4,

5,

a,

a + d,

a + 2d,

a + 3d,

a + 4d, etc.

The coefficient of d is equal to the number of terms preceding the one in which it occupies a place. Thus the fifth term is a + 4d, in which the coefficient 4 equals the number of the preceding terms.

From this we learn the rule by which at once to designate any term without finding all the preceding terms. For the one hundredth term we should have a + 99 d, or, if the number of terms be represented by n, then the last term would be represented by -

l = a + (n - 1) d. (119.)

For example, in a progression where a, the first term, equals 1, d the difference, 2, and n, the number of terms, 90, the last term will be -

l = a + (n - 1) d = 1 + (90 - 1) 2 = 179.

Therefore, to find the last term:

To the first term add the product of the common difference into the number of terms less one.

By a transposition of the terms in the above expression, so as to give it this form -

a = l - (n - 1) dy (120.)

we have a rule by which to find the first term, which, in words, is -

Multiply the number of terms less one by the common difference, and deduct the product from the last term; the remainder will be the first term.

By a transposition of the terms of the former expression to this form -

l - a = (n - 1) d,

and dividing both members by (n - 1), we have -

d = 1-a/n-1; (121.)

which is a rule for the common difference, and which, in words, is -

Subtract the first term from the last, and divide the remainder by the number of terms less one; the quotient will be the common difference.

Multiplying both members of the equation (121.) by (n - 1) and dividing by d, we obtain -

n-1 = l-a/d

Transferring 1 to the second member, we have -

n = l-a/d + 1; (122.)

which is a rule for finding the number of terms, and which, in words, is -

Divide the difference between the first and last terms by the common difference; to the quotient add unity, and the sum will be the number of terms.

Thus it has been shown, in equations (119,) (120), (121), and (122), that when, of the four quantities in arithmetical progression, any three are given, the fourth may be found.

The sum of the terms of an arithmetical progression may be ascertained by adding them; but it may also be had by a shorter process. If the terms are written in order in a horizontal line, and then repeated in another horizontal line beneath the first, but in reversed order, as follows:

1, 3, 5, 7, 9, 11, 13, 15,

15,13, 11, 9, 7, 5, 3, 1,

16, 16, 16, 16, 16, 16, 16, 16,

and the vertical columns added, the sums will be equal. In this case the sum of each vertical couple is 16, and there are 8 couples; hence the sum of these 8 couples is 8 x 16 = 128. And in general the sum will be the product of one of the couples into the number of couples. It will be observed that the first couple contains the first and last terms, 1 and 15; therefore the sum of the double series is equal to the product of the sum of the first and last terms into the number of terms. Or if 5 be put to represent the sum of the series, we shall have -

2 S = (a + l) n,

and, dividing both sides by 2 -

S=(a +l)n/2; (123.)

Or, in words: The sum of an arithmetical series equals the product of the sum of the first and last terms, into half the number of terms.