that is, the coefficient of d in any term is always less by 1 than the number which marks the place of the term. In other words, to find any term, we multiply the common difference by a number less by 1 than that which marks the place of the term, and add the product to the first term when the progression is increasing, but subtract the product from the first term when the progression is decreasing.

Hence if, in addition to our previous notation, we denote the number of terms by n, and the last term by 7, we have the formula

l=a+ (n-1) d, in an increasing progression; and

la-(n-1) d, in a decreasing progression. If the double sign be used, the general formula for the last term is

l=a(n-1) d. Hence,

To find the last term, multiply the common difference by the number of terms minus one, and add the product to the first term if the progression is increasing, but subtract the product from the first term if the progression is diminishing.

1. Required the 12th term of the progression, 7, 10, 13, 16, &c.

In this example, a=7, d= 3, and n=12; and by substituting these numbers in the formula, 7=a+(n−1) d, we have

7=7+(12—1)3=7+11.3=7+33=40.

Therefore, the 12th or last term is 40.

2. Required the 9th term of 60, 55, 50, &c.

In this example, a= = 60, d=5, and n=9, and the progression is decreasing. Hence, l—a—(n-1) d becomes, by substitution,

7=60-(9-1) 560-8.5=60-40=20.

ART. 136. Let us now proceed to find a formula for the sum of any number of terms. For this purpose, let S represent the sum of n terms of the progression, a, a+d, a +2d, &c. Then,

1st

2d

3d

4th

nth

s=a+(a+d)+(a+2d)+(a+3d)+......+7. (1).

If we write the progression in the reverse order, beginning with the last term, it is plain that the successive terms of the same progression will be 1, 1-d, 1—2d, &c. Hence,

nth

(n-1)st

(n-2)d

(n-3)d

1st

S=1+(i−d)+(l−2 d)+(l−3d)+. +a. (2).

......

Remark. It is manifest that the terms cannot all be written, unless some determinate value is given to n. We therefore use points to supply the place of the indefinite number of terms.

By adding equations (1) and (2), and observing that d, 2d, 3d, &c. in (1), are cancelled byd, — 2d, -3d, &c. in (2), we have

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

But since, in this last equation, the quantities included. between the several parentheses are the same, and since this same quantity a+ is repeated as many times as there are terms in the progression, that is, n times, the second member is the same as n (a+1). Hence,

To find the sum of any number of terms in progression by difference, multiply the sum of the first and last terms by half the number of terms, or multiply half the sum of the first and last terms by the number of terms.

Required the sum of 8 terms of the series 6, 10, 14, &c. In this example, a = 6, d= 4, and n8. We are first to find the value of 1, which by the preceding Article is 7-6(8-1)4=34. Then, substituting the values of a, l, and n in the formula for S, we have

S=(6+34)=4.40 160.

=

SECTION XLIII.

EXAMPLES IN PROGRESSION BY DIFFERENCE.

ART. 137. 1. Required the 12th term of the series 10, 16, 22, &c.

2. Required the 20th term of the series 100, 98, 96, &c.

3. What is the sum of 100 terms of 1, 2, 3, 4, &c.? 4. Find the 8th term and the sum of the first 8 terms of 7, 10, 13, &c.

5. Required the sum of 10 terms of the series, in which the first term is 2, and the common difference. 6. Required the 25th term, and the sum of the first 25 terms of the series 60, 592, 59, &c.

7. A man buys 10 sheep, giving 2 s. for the first, 4 s. for the second, 6 s. for the third, and so on. How much do they all cost him?

8. Twenty stones and a basket are in the same straight line, and 5 yards asunder; how far would a boy travel, if, starting from the basket, he were to pick up the stones, and carry them one by one to the basket?

9. Separate 39 into three parts which shall be in arithmetical progression, the common difference being 7.

Let x= the least part, or first term of the progression.

10. Find three numbers in arithmetical progression such that their sum shall be 30, and their continued product 750.

Let y= the common difference, and x-the middle term. Then x -y, x, and x+y will represent the

numbers.

11. Two men, 189 miles asunder, set out at the same time to travel towards each other till they meet. One of them goes 10 miles each day; the other goes 3 miles the first day, 5 the second, 7 the third, and so on. In how many days will they meet?

Let x the number of days; then x will represent the number of terms, and will correspond to n in the formula.

12. Two travellers, 135 miles asunder, set out at the same time to travel towards each other. One travels 5 miles the first day, 8 the second, 11 the third, and so on; the other travels 20 miles the first day, 18 the second, 16 the third, and so on. In how many days will they meet?

SECTION XLIV.

PROGRESSION BY QUOTIENT.

ART. 138. A progression by quotient, or geometrical progression, is a series of quantities such, that if any one of them be divided by that which immediately precedes it, the quotient will be the same, in whatever part of the series the two quantities are taken. The successive quantities are called terms of the progression.

The quotient arising from the division of any term by

that which immediately precedes it, is called the common ratio.

For example, 3, 6, 12, 24, &c. is a progression by quotient, the common ratio being 2; also, 100, 20, 4, ,, &c. is a similar progression, the common ratio being.

A progression by quotient is called increasing or decreasing, according as the terms increase or diminish from left to right. The former of the preceding progressions is increasing, the latter decreasing.

ART. 139. In order to exhibit a progression by quotient in its most general form, let a, b, c, d, &c. represent the successive terms at the commencement of the series, and let q be the common ratio.

Now, since from the definition of a progression by quotient, each term is equal to q times the preceding term we have

b=aq, c=aq2, d= a q3, e= a qa, &c.

Representing the last term by 7, and supplying the place of the indefinite number of intermediate terms by dots, the terms of the progression will be

We see that the exponent of q, in any term of this series, is less by 1 than the number which marks the place of the term. Thus, the 5th term is a q4, the 6th is a q5, &c. Hence, if n represent the number of terms, the nth or last term will be a q"-1. But I also represents the last term. Therefore,

This is the formula for the last term. Hence,

To find any term of a progression by quotient, multiply the first term by that power of the common ratio, denoted