The numbers which form the series, are called the terms of the progression.

Note. The first and last terms of a progression are called the extremes, and the other terms the means.

Any three of the five following things being given, the other two may be easily found.

1.

The first term.

2. The last term.

3. The number of terms.

4. The common difference.
5. The sum of all the terms.

PROBLEM I.

The first term, the last term, and the number of terms being given, to find the common difference.

RULE.*

Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference sought.

EXAMPLES.

1st. The extremes are 3 and 39, and the number of terms is 19: What is the common difference?

39

3

36

Extremes.

Divide by the number of terms less 1=19-1=18)36(2 Ans.

28. A man had 10 sons, whose several ages differed alike; the youngest was 3 years old, and the eldest 48; What was the common difference of their ages?

Ans. 5.

3d. A man is to travel from Boston to a certain place in 9 days, and to go but 5 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 37 miles: Required the daily increase?

PROBLEM II.

Ans. 4.

The first term, the last term, and the number of terms being given, to find the sum of all the terms.

RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the answer.

* The difference of the first and last terms evidently shews the increase of the first term by all the subsequent additions, till it becomes equal to the last; and as the number of those additions was one less than the number of terms, and the increase, by every addition, equal, it is plain that the total increase, divided by the number of additions, must give the difference of every one separately; whence the rule is manifest.

+ Suppose another series of the same kind with the given one be placed under it in an inverse order; then will the sum of any two corresponding terms be the

EXAMPLES.

1st. The extremes of an arithmetical series are 3 and 39, and the number of terms 19: Required the sum of the series?

24. It is required to find how many strokes the hammer of a clock would strike in a week, or 168 hours, provided it increased 1 at each hour?

Ans. 14196.

3d. Suppose a number of stones were laid a yard distant from each other for the space of a mile, and the first a yard from a basket: What length of ground will that man travel over, who gathers them up singly, returning with them one by one to the basket? Ans. 1761 miles.

N. B. In this question, there being 1760 yards in a mile, and the man returning with each stone to the basket, his travel will be doubled; therefore the first term will be 2, and the last 1760×2, and the number of terms 1760.

4th. A man bought 25 yards of linen in Arithmetical Progression; for the 4th yard he gave 12 cents, and for the last yard 75 cents: What did the whole amount to, and what did it average per yard?

3 the common difference by which the first term is found

[to be 3.

75+3×25

Then

$9 75c. and the average price is 39cts. per yard.

2

5th. Required the sum of the first 1000 numbers in ther natural order? Ans. 500500.

same as that of the first and last; consequently, any one of those sums, multiplied by the number of terms, must give the whole sum of the two series.

Let 1, 2, 3, 4, 5, 6, 7, 8, be the given series.
And 8, 7, 6, 5, 4, 3, 2, 1, the same inverted.
Then, 9+9+9+9+9+9+9+9=9×8=72, and

72

1+2+3+4+5+6+7+3=-=36.

PROBLEM III.

Given the extremes and the common difference, to find the number of

terms.

RULE.* Divide the difference of the extremes by the common difference, and the quotient increased by 1 will be the number of terms required.

EXAMPLES.

1st. The extremes are 3 and 39, and the common difference ?: What is the number of terms?

2J. A man going a journey, travelled the first day 7 miles, the last day 51 miles, and cach day increased his journey by 4 miles How many days did he travel, and how far?

Ans. 12 days, and 348 miles.

PROBLEM IV.

The extremes and common difference given, to find the sum of ali

the series.

RULE. Multiply the sum of the extremes by their difference increased by the common difference, and the product divided by twice the common difference will give the sum.†

EXAMPLES.

1st. If the extremes are 3 and 39, and the common difference 2 : What is the sum of the series?

By the first Problem, the difference of the extremes, divided by the number of terms less 1, gave the common difference; consequently the same divided by the common difference, must give the number of terms less 1; hence, this quotient, augmented by 1, must be the answer to the question.

+ By the 3d Problem find the number of terms, and then, with the number of terms and the extremes, find, by Prob. 2, the sum of the series. This is the rule, which is contracted in the text. Thus in the 1st Example, by Problem 3, 39-3 +1 the number of terms, and by Prob. 2, 39+3x39-3+1 twice the sum of 2

39+3=42 sum of the extremes.

39-3=36 difference of extremes.

36+2=38=difference of extremes increased by the common

2X2

=399.

399

2d. A owes B a certain sum, to be discharged in a year, by pay ing 6d. the first week, 18d. the second, and thus to increase every weekly payment by a shilling, till the last payment be 21. 11s. 6d. : Ans. £67 123. What is the debt?

PROBLEM V.

The extremes and sum of the series given, to find the number of terms

RULE.

Twice the sum of the series, divided by the sum of the extremes, will give the number of terms.*

EXAMPLES.

1st. Let the extremes be 3 and 39, and the sum of the series 339: What is the number of terms?

Sum of the series=399

Sum of the extremes=39+3=42)798(19 Ans.

42

378

978

2d. A owes B 671. 123. to be paid weekly in Arithmetical Progression, the first payment to be 6d. and the last to be 51s. 6d. : How many payments will there be, and how long will he be in discharging the debt?

Ans. 52 payments, and as many weeks.

This Problem is the reverse of Prob. II. and the reason of the rule is obvious from the demonstration of the Rule, Prob. IL

PROBLEM VI.

The extremes and the sum of the series given, to find the common

difference.
RULE.

Divide the product of the sum and difference of the extremes, by the difference of twice the sum of the series, and the sum of the extremes, and the quotient will be the common difference.* EXAMPLES.

1st. Let the extremes be 3 and 39, and the sum 399: What is the common difference?

Sum of the extremes = 39+ 3 = 42

Diff. of the extremes

39-3X 36

399×2-39+3

Or,

2d. A owes B £67 12s. to be discharged in a year, by weekly payments; the first payment to be 6d. and the last, £2 11s 6d. : What is the common difference of the payments, and what will each payment be?

The first term, the common difference, and the number of terms given, to find the last term.

RULE.

The number of terms less 1, multiplied by the common difference, and the first term added to the product, will give the last term.t

EXAMPLES.

1st. If the first term be 3, the common difference 2, and the number of terms 19: What is the last term?

This rule is only a contraction of the following process. By Prob. V. find the number of terms, and, then, from the extremes and number of terms, find by Prob. I. the common difference.

By Prob. I. the difference of the last and first terms divided by the number of terms less 1, gives the common difference, whence the common difference multiplied by the number of terms less 1, and the product increased by the first. term, must give the last term.

E e