ARITHMETICAL PROGRESSION.

291. An Arithmetical Progression is a series of quantities which increase or decrease by a common difference.

Thus, 1, 3, 5, 7, 9, etc., or 12, 9, 6, 3, etc., and a, a+d, a+2d, etc., a, a-d, a-2d, etc., are in Arithmetical Progression.

The series is said to be increasing or decreasing, according as d is positive or negative.

292. To investigate a rule for finding any term of an arithmetical progression, take the following series, in which the first line denotes the number of each term, the second an increasing arithmetical series, and the third a decreasing arithmetical series.

It is manifest that the coefficient of d in any term is less by unity than the number of that term in the series; therefore, the nth term =a+(n−1)d.

If we designate the nth term by 7, we have

l=a+(n−1)d, when the series is increasing, and
l—a—(n−1)d, when the series is decreasing. Hence,

Rule for finding Any Term of an Arithmetical Series.Multiply the common difference by the number of terms less one; when the series is increasing, add this product to the first term; when decreasing, subtract it from the first term.

The equation 7=a+(n−1)d, contains four variable quantities, any one of which may be found when the other three are known.

293. Having 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 any arithmetical series, as the following, and write the same series under it in an inverted order, we have

Adding,

Whence,

S=1+3 + 5+ 7+9+11,
S=11+9 + 7+ 5+ 3+ 1.

28=12+12+12+12+12+12.

28 12X the number of terms, =12×6=72.
S= of 72=36, the sum of the series.

To render this method general, let 7 the last term, and write the series both in a direct and inverted order.

Then, Sa+(a+d)+(a+2d)+(a+3d). .

And, Sl(l—d) + (l—2d) + (l—3d). . +a.

28=(1+a)+(l+a)+(l+a)+(l+a)... +(l+a),
28=(1+a) taken as many times as there are terms (n) in

the series. Hence,

28=(1+a)n;

s=(1+a)n =(1+a)n. Hence,

Rule for finding the Sum of an Arithmetical Series.Multiply half the sum of the two extremes by the number of

terms.

It also appears that

The sum of the extremes is equal to the sum of any other two terms equally distant from the extremes.

n

294. The equations 7=a+(n−1)d, and S=(a+1), furnish the means of solving this general problem:

Knowing any three of the five quantities, a, d, l, n, S, which enter into an arithmetical series, to determine the other

two.

The following table contains the results of the solution of all the different cases. As, however, it is not possible to retain these in

the memory, it is best, in ordinary cases, to solve all examples in Arithmetical Progression by the above two formulæ :

1. Find the 15th term of the series 3, 7, 11,

etc. Ans. 59.

Here, a=3, n—1=14, and d=4. Substituting these values in formula (1), we have 7=3+14×4=3+56=59.

etc.

2. Find the 20th term of the series 5, 1, 3, etc. Ans. -71.

3. Find the 8th term of the series,,, etc.

Ans.

4. Find the 30th term of the series -27,-20, -13, Ans. 176.

5. Find the nth term of 1+3+5+7. Ans. 2n-1.

6. Find the sum of 1+2+3+4, etc., to 50 terms.

From formula (1), we find l=50. Substituting this in formula (2), we have S=(1+50)25=1275, Ans. Or, use formula 5.

d = {/ 7.08 7+38 +15+, etc., to 16 terms.

8. Of 12+8+4+, etc., to 20 terms.

9. Of 2+2}+2}+, etc., to n terms.

Ans. 142.

n

10. Of 1———, etc., to n terms. A. 12(18—7n).

12. If a falling body descends 16 feet the 1st sec., 3 times this distance the next, 5 times the next, and so on, how far will it fall the 30th sec., and how far altogether in half a min,? Ans. 94811, and 14475 ft.

13. Two hundred stones being placed on the ground in a straight line, at the distance of 2 feet from each other;

258

RAY'S ALGEBRA, SECOND BOOK.

how far will a person travel who shall bring them separately to a basket, which is placed 20 yards from the first stone, if he starts from the spot where the basket stands? Ans. 19 miles, 4 fur., 640 ft.

14. Insert 3 arithmetical means between 2 and 14.

¦ Here, a=2, 7=14, and n=5. From formula (1), we obtain d=3. Hence, the three means will be 5, 8, and 11.

To solve this problem generally, let it be required to insert m arithmetical means between a and 7.

Since there are m terms between a and 7, we shall have n=m+2, and formula (1) becomes la+(m+1)d. Hence, d=Therefore,

m+1

The common difference will be equal to the difference of the extremes divided by the number of means plus one.

15. Insert 4 arithmetical means between 3 and 18. Ans. 6, 9, 12, 15.

16. Insert 9 arithmetical means between 1 and -1. Ans.,, etc., to

17. How many terms of the series 19, 17, 15, etc., amount to 91? Ans. 13, or 7. From (2) and (1), find n, or use formula 14. Explain this result.

18. How many terms of the series .034, .0344, .0348, etc., amount to 2.748? Ans. 60.

19. The sum of the first two terms of an arithmetical progression is 4, and the fifth term is 9; find the series. Ans. 1, 3, 5, 7, 9, etc.

20. The first two terms of an arithmetical progression being together =18, and the next three terms =12, how many terms must be taken to make 28? Ans. 4, or 7.

21. In the series 1, 3, 5, etc., the sum of 2r terms: the sum of terms::x: 1; determine the value of x.

Ans. 4.

2.

44