348. To investigate the properties of an arithmetical progression, we may suppose the series to terminate; there will then be five parts or elements: the first term, the last term, the number of terms, the common difference, and the sum of the terms. The first term and last term are called the extremes, and all the terms between the extremes are called arithmetical means.

349. In an Arithmetical Progression, the last term is equal to the first term plus the common difference multiplied by the number of terms less 1.

Let a denote the first term, 7 the last term, d the common difference, and n the number of terms; then the series will be represented thus:

a, (a + d), (a + 2d), (a + 3d), . . . . l.

And we perceive that in every term the coefficient of d is equal to the number of preceding terms; hence,

1 = a + (n − 1) d . . . (4),

in which is positive or negative, according as the series is an increasing or a decreasing one.

350. In an arithmetical progression the sum of any two terms equidistant from the extremes is equal to the sum of the extremes.

Let t denote a term of the series which has r terms before it, and t'a term which has r terms after it; then the terms, t and ť, will be equidistant from the extremes. Suppose the series to be increasing; then from the nature of the series,

351. The sum of the terms of an Arithmetical Progression is equal to one half the sum of the two extremes, multiplied by the number of terms.

Represent the sum of the series by S; then we have

S = a + (a + d) + (a + 2d) +

....

By writing the series in a reversed order, we have also
S=1+ (1 - d) + (1 − 2d) + ... + a

(1).

(2).

2S

Therefore, by addition,

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

(3).

Now equation (3) expresses the sum of n terms, each equal to (a+1); hence,

2S = n (a + 1);

352. To insert any number of arithmetical means between two given terms.

Let n' denote the number of means to be inserted. Then the number of terms in the completed series will be n' + 2; and we shall have n = n' + 2.

This value of n substituted in formula (4) (349), gives

Having the common difference, the means are readily obtained.

contain, in all, five quantities, a, l, n, d, S, four of which enter each equation. Now if any three of these quantities be given, the other two may be found; for, if the values of the three given quantities be substituted in the formulas, there will result two equations containing only two unknown quantities.

1. The first term of an arithmetical series is 5, the common difference 3, and the number of terms 24. Find the last term, and the sum of the series.

2. Given a = 15, d

terms.

= 2, and S= 60, to find the number of

Substituting the given values in (A) and (B), we have

Both values of n are possible; for there are two series answering to the given conditions, one having 6 terms, and the other 10; these are

and

4

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

15, 13, 11, 9, 7, 5.

The sum of either series is 60.

1. The first term of an arithmetical series is 7, the common difference 3, and the number of terms 36; find the last term.

Ans. 112.

2. The first term of an arithmetical series is 275, the last term 5, and the number of terms 46; required the sum of the terms. Ans. 6440.

3. The sum of an arithmetical series is 156, the number of terms 8, and the common difference 5. Required the extremes. 4. Find the sum of the terms in an arithmetical progression, knowing that the first term is 1, the common difference, and the number of terms 101. Ans. 2626.

5. Find four arithmetical means between 7 and 37.

Ans. 13, 19, 25, 31.

6. The first term of an arithmetical series is 3, the number of terms 60, and the sum of the terms 3720; required the common difference, and the last term. Ans. d = 2, 1 = 121.

7. What will be the sum of the series if 9 arithmetical means be inserted between 9 and 109? Ans. 649.

8. If three arithmetical means be inserted between and 1, what will be the common difference? Ans. 4

9. What debt can be discharged in a year by paying 1 cent the first day, 3 cents the second, 5 cents the third, and so on, increasing the payment each day by 2 cents? Ans. $1332.25.

10. A footman travels the first day 20 miles, 23 the second, 26 the third, and so on, increasing the distance each day 3 miles. How many days must he travel at this rate to go 438 miles?

Ans. 12.

11. Find the sum of n terms of the progression 1, 2, 3, 4, Ans. S=22 (1+ n).

5, 6,

.....

2

12. Find the sum of n terms of the progression 1, 3, 5, 7, ....

S = n2.

بیل

13. The sum of the terms of an arithmetical series is 950, the common difference is 3, and the number of terms 25. the first term?

What is Ans. 2.

14. A man bought a certain number of acres of land, paying for the first $; for the second, $3; and so on. When he came to settle he had to pay $3775. How many acres did he purchase? Ans. 150 acres.

15. The 14th, 134th, and last terms of an arithmetical progression are 66, 666, and 6666, respectively. Required the number of terms.

Ans. 1334.

THE TEN CASES.

354. Given any three of the quantities, a, 7, n, d, 8, to find the other two.

This problem will present ten cases, each giving rise to two formulas, making in all twenty different formulas, or four values for each letter. The results in each case may be obtained directly from the two fundamental equations, or those of any particular case may be derived from some preceding case, as is most convenient. The whole will be left as an exercise for the student.