and the last 17 miles, and increases each day's travel by an equal difference, what is the daily increase?

Ans. 8 miles. ART. 305. To find the sum of all the terms, the first term, last term, and number of terms, being given.

ILLUSTRATION.

follows:

2, 5, 8, 11,

20, 17, 14, 11,

22, 22, 22, 22,

22, 22, 22, = 154, sum of both series.

From the arrangement of the above series, we see that, by adding the two as they stand, we have the same number for the sum of the successive terms, and that the sum of both series is double the sum of either series.

It is evident that, if 22 in the above series be multiplied by 7, the number of terms, the product will be the sum of both series; thus, 22 × 7: 154; and, therefore, the sum of either series will be 1542=77. But 22 is the sum of the extremes in each series; thus, 20 +2=22. Therefore, if the sum of the extremes be multiplied by the number of terms, the product will be double the sum of either series. Hence,

RULE 1.- Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series. Or,

RULE 2. Multiply the sum of the extremes by half the number of terms, and the product is the sum required.

EXAMPLES FOR PRACTICE.

1. If the extremes of a series are 5 and 45, and the number of terms 9, what is the sum of the series?

(45+5)×9
2

OPERATION.

Ans. 225.

= 225, sum of the series.

2. John Oaks engaged to labor for me 12 months. For the first month I was to pay him $7, and for the last month $51. In each successive month he was to have an equal addition to his wages; what sum did he receive for his year's labor?

Ans. $348.

QUESTION. - -Art. 305. What is the rule for finding the sum of all the terms, the first term, last term, and number of terms, being given?

3. I have purchased from W. Hall's nursery 100 fruit-trees of various kinds, to be set around a circular lot of land, at the distance of one rod from each other. Having deposited them on one side of the lot, how far shall I have travelled when I have set out my last tree, provided I take only one tree at a time, and travel on the same line each way? Ans. 9801 rods.

ART. 306. To find the number of terms, the extremes and common difference being given.

ILLUSTRATION. Let the extremes of a series be 2 and 29, and the common difference 3. The difference of the extremes will be 29 — 2 = 27. Now, it is evident that, if the difference of the extremes be divided by the common difference, the quotient will be the number of common differences; thus, 27 ÷ 3 = 9. It has been shown (Art. 303) that the number of terms is 1 more than the number of differences; therefore, 9+1 10 is the number of terms in this series. Hence the following

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 FOR PRACTICE.

1. If the extremes of a series are 4 and 44, and the common difference 5, what is the number of terms?

Ans. 9.

2. A man going a journey travelled the first day 8 miles, and the last day 47 miles, and each day increased his journey by 3 miles. How many days did he travel? Ans. 14 days.

ART. 307. To find the sum of the series, the extremes and common difference being given.

ILLUSTRATION.-Let the extremes be 2 and 29, and the common difference 3. The difference of the extremes will be 29 2=27; and it has been shown (Art. 306) that if the difference of the extremes be divided by the common difference, the

QUESTION.-Art. 306. What is the rule for finding the number of terms, the extremes and common difference being given?

quotient will be the number of terms less one. Therefore, the number of terms less one will be 27 ÷ 3 = 9, and the number of terms 9+1 = = 10. It was also shown (Art. 305) that, if the number of terms be multiplied by the sum of the extremes, and the product divided by 2, the quotient will be the sum of the series. Hence the

RULE. Divide the difference of the extremes by the common difference, and to the quotient add 1; multiply this sum by the sum of the extremes, and half the product is the sum of the series.

EXAMPLES FOR PRACTICE.

1. If the two extremes are 11 and 74, and the common difference 7, what is the sum of the series?

Ans. 425.

= 425, sum of series.

2. A pupil commenced Virgil by reading 12 lines the first day, 17 lines the second day, and thus increased every day by 5 lines, until he read 137 lines in a day. How many lines did he read in all ? Ans. 1937 lines. ART. 308. To find the last term, the first term, the number of terms, and the common difference, being given.

ILLUSTRATION. Let the first term of a series be 2, the number of terms 10, and the common difference 3. It has been shown (Art. 304) that the number of common differences is always 1 less than the number of terms; and that the sum of the common differences is equal to the difference of the extremes; therefore, since the number of terms is 10, and the common difference 3, the difference of the extremes will be (10 — 1) × 3 27; and this difference, added to the first term, must give the last term; thus, 2+27 =29. Hence the following

RULE.-Multiply the number of terms less 1 by the common difference, and add this product to the first term for the last term.

NOTE. If the series is descending, the product must be subtracted from the first term.

QUESTIONS. Art. 307. What is the rule for finding the sum of the series, the extremes and common difference being given ?-Art. 308. What is the rule for finding the last term, the first term, the number of terms, and common difference, being given?

EXAMPLES FOR PRACTICE.

1. If the first term is 1, the number of terms 7, and the comnon difference 6, what is the last term?

Ans. 37.

2. If a man travel 7 miles the first day of his journey, and 9 miles the second, and shall each day travel 2 miles further than the preceding, how far will he travel the twelfth day?

Ans. 29 miles.

3. If A set out from Portland for Boston, and travel 201 miles the first day, and on each succeeding day 1 miles less than on the preceding, how far will he travel the tenth day?

Ans. 63 miles.

ANNUITIES AT SIMPLE INTEREST BY ARITHMETICAL PRO

GRESSION.

ART. 309. AN ANNUITY is a sum of money to be paid annually, or at any other regular period, either for a limited time or forever.

The present worth of an annuity is that sum which being put at interest will be sufficient to pay the annuity.

The amount of an annuity is the interest of all the payments added to their sum.

Annuities are said to be in arrears when they remain unpaid after they have become due.

ART. 310. To find the amount of an annuity at simple interest.

Ex. 1. A man purchased a farm for $2000, and agreed to pay for it in 5 years, paying $400 annually; but, finding himself unable to make the annual payments, he agreed to pay the whole amount at the end of the 5 years, with the simple interest, at 6 per cent., on each payment, from the time it became due till the time of settlement; what did the farm cost him?

ILLUSTRATION.

Ans. $2240.

It is evident the fifth payment will be $400, without interest; the fourth will be on interest 1 year, and will amount to $424; the third will be on interest 2 years, and will amount to $448; the second will be on interest 3

QUESTIONS. Art. 309. What is an annuity? What is meant by the present worth of an annuity? By the amount? When are annuities said to be in arrears?

years, and will amount to $472; and the first will be on interest 4 years, and will amount to $496. Therefore, these several sums form an arithmetical series; thus, 400, 424, 448, 472, 496; of which the fifth payment, or the annuity, is the first term, the interest on the annuity for one year the common difference, the time in years, the number of terms, and the amount of the annuity, the sum of the series. The sum of this series is found by (400+496) × 5 = $2240. Hence the 2

Art. 305; thus,

RULE.

- First find the last term of the series (Art. 308), and then the sum of the series (Art. 305).

NOTE.If the payments are to be made semi-annually, quarterly, &c., these periods will be the number of terms, and the interest of the annuity for each period the common difference.

EXAMPLES FOR PRACTICE.

2. What will an annuity of $250 amount to in 6 years, at 6 per cent. simple interest? Ans. $1725. 3. What will an annuity of $380 amount to in 10 years, at 5 per cent. simple interest? Ans. $4655.

4. An annuity of $825 was settled on a gentleman, January 1, 1840, to be paid annually. It was not paid until January 1, 1848; how much did he receive, allowing 6 per cent. simple interest? Ans. $7986.

5. A gentleman let a house for 3 years, at $200 a year, the rent to be paid semi-annually, at 8 per cent. per annum, simple interest. The rent, however, remained unpaid until the end of the three years; what did he then receive? Ans. $660.

6. A certain clergyman was to receive a salary of $700, to be paid annually; but, for certain reasons, which we fear were not very good, his parishioners neglected to pay him for 8 years; but he agreed to settle with them, and allow them $100 if they would pay him his just due with interest; required the sum received. Ans. $6676.

7. A certain gentleman in Boston has a very fine house, which he rents at $50 per month. Now, if his tenant shall omit payment until the end of the year, what sum should the owner receive, reckoning interest at 12 per cent.? Ans. $633.

QUESTIONS. Art. 310. What forms the first term of a progression in an annuity? What the common difference? What the number of terms? What the sum of the series? What is the rule for finding the amount of an annuity at simple interest? If the payments are made semi-annually, quarterly, &c., what constitute the terms? What the common difference?