9. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for the first yard, and 301 cents for the last yard; what was the average price per yard, and what was the amount of the whole?

Since the price of each succeeding yard increases by a constant excess, it is plain, the average price is as much less than the price of the last yard, as it is greater than the price of the first yard; therefore, one half the sum of the first and last price is the average price.

One half of 4 cts. +301 cts. =1521⁄2 cts. — average price; and the price, 152 cts. X 100 15250 cts. $152'50, whole cost.

Ans.

Hence, when the extremes and the number of terms are given, to find the sum of all the terms,-Multiply the sum of the extremes by the number of terms, and the product will be the answer.

Ans. 46055.

10. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series? 11. What is the sum of the first 100 natural order, that is, 1, 2, 3, 4, &c. ? 12. How many times does a common clock strike in 12 hours?

numbers, in their Ans. 5050.

Ans. 78.

13. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money?

The last year's rent will evidently be $50 without interest, the last but one will be the amount of $50 for 1 year, the last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years $107.

If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series? Ans. $1570. 14. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent. simple interest? Ans. $7900.

15. There are, in a certain triangular field, 41 rows of corn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how many hills of corn in the field? Ans. 1681 hills.

16. If a triangular piece of land, 30 rods in length, be 20 rods wide at one end, and come to a point at the other, what number of square rods does it contain? Ans. 300. 17. A debt is to be discharged at 11 several payments, in arithmetical series, the first to be $5, and the last $75; what is the whole debt? the common difference between the several payments?

Ans. whole debt, $440; common difference, $7. 18. What is the sum of the series 1, 3, 5, 7, 9, &c., to 1001 ? Ans. 251001.

Note. By the reverse of the rule under ex. 5, the difference of the extremes 1000, divided by the common difference 2, gives a quotient, which, increased by 1, is the number of

terms

501.

19. What is the sum of the arithmetical series 2, 21, 3, 31, 4, 41, &c., to the 50th term inclusive ? Ans. 7121. 20. What is the sum of the decreasing series 30, 29, 291, 29, 28, &c., down to 0?

Note. 30+1= 91, number of terms. Ans. 1365.

QUESTIONS.

1. What is an arithmetical progression? 2. When is the series called ascending? 3. when descending? 4. What are the numbers, forming the progression, called? 5. What are the first and last terms called? 6. What are the other terms called? 7. When the first term, common difference, and number of terms, are given, how do you find the last term? 8. How may arithmetical progression be applied to simple interest? 9. When the extremes and number of terms are given, how do you find the common difference? how do you find the sum of all the terms?

10.

113. Any series of numbers, continually increasing by a constant multiplier, or decreasing by a constant divisor, is called a Geometrical Progression. Thus, 1, 2, 4, 8, 16, &c. is an increasing geometrical series, and 8, 4, 2, 1, , t, &c. is a decreasing geometrical series.

As in arithmetical, so also in geometrical progression, there are five things, any three of which being given, the ather two may be found :

1st. The first term.

28. The last term.

3d. The number of terms.

4th. The ratio.

5th. The sum of all the terms.

The ratio is the multiplier or divisor, by which the series is formed.

1. A man bought a piece of silk, measuring 17 yards, and, by agreement, was to give what the last yard would come to, reckoning 3 cents for the first yard, 6 cents for the second, and so on, doubling the price to the last; what did the piece of silk cost him?

3 X 2 X 2 X 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 196608 cents, $1966'08, Answer,

In examining the process by which the last term (196608) has been obtained, we see, that it is a product, of which the ratio (2) is sixteen times a factor, that is, one time less than the number of terms. The last term, then, is the sixteenth power of the ratio, (2,) multiplied by the first term (3.)

Now, to raise 2 to the 16th power, we need not produce all the intermediate powers; for 24 2 X 2 × 2 × 2 = 16, is a product of which the ratio 2 is 4 times a factor; now, if 16 be multiplied by 16, the product, 256, evidently contains the same factor (2) 4 times + 4 times, 8 times; and 256 X 256 65536, a product of which the ratio (2) is 8 times 8 times, 16 times, factor; it is, therefore, the 16th power of 2, and, multiplied by 3, the first term, gives 196608, the last term, as before. Hence,

When the first term, ratio, and number of terms, are given, to find the last term,——

I. Write down a few leading powers of the ratio with their indices over them.

II. Add together the most convenient indices, to make an index less by one than the number of the term sought.

III. Multiply together the powers belonging to those indices, and their product, multiplied by the first term, will be the term sought.

2. If the first term be 5, and the ratio 3, what is the 8th term?

Powers of the ratio, with their indices over them.

I 2 3 + 4 =

3, 9, 27, × 81 ≤ 2187 × 5 first term, 10935, Answer.

3. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels; these he plants the second year; now, supposing the annual increase to continue 8 fold, what would be the produce of the 16th year, allowing 1000 kernels to a pint ? Ans. 2199023255'552 bushels.

4. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years?

217 131072.

Ans. 1310 72.

5. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 cents for the second, and so on, in 3 fold ratio; what did the whole cost him?

2+6+18+54= 80 cents.

Ans. 80 cents.

In a long series, the process of adding in this manner would be tedious. Let us try, therefore, to devise some shorter method of coming to the same result. If all the terms, excepting the last, viz. 2+6+18, be multiplied by the ratio, 3, the product will be the series 6+18+ 54; subtracting the former series from the latter, we have, for the remainder, 54—2, that is, the last term, less the first term, which is evidently as many times the first series (2+6+18) as is expressed by the ratio, less 1: hence, if we divide the difference of the extremes (54 - 2) by the ratio, less 1, (31,) the quotient will be the sum of all the terms, excepting the last, and, adding the last term, we shall have the whole amount. Thus, 54 - 2 - 52, and 3-1 = 2; then, 522 26, and 54 added, makes 80, Answer, as before. Hence, when the extremes and ratio are given, to find the sum of the series,—Divide the difference of the extremes by the ratio, less 1, and the quotient, increased by the greater term, will be the answer.

6. If the extremes be 4 and 131072, and the ratio 8, what is the whole amount of the series?

131072-4

+131072149796 Answer.

8 1

7. What is the sum of the descending series 3, 1, 4, t, ,&c., extended to infinity?

It is evident the last term must become 0, or indefinitely near to nothing; therefore, the extremes are 3 and 0, and Ans. 4.

the ratio 3.

8. What is the value of the infinite series 1+1+16+ 64, &c. ?

Ans. 13.

9. What is the value of the infinite series, O + 180 + robo + Todoo, &c., or, what is the same, the decimal '11111, &c., continually repeated? Ans.

10. What is the value of the infinite series, T0 + TOBOO, &c., descending by the ratio 100, or, which is the same, the repeating decimal '020202, &c.? Ans.

11. A gentleman, whose daughter was married on a new year's day, gave her a dollar, promising to triple it on the first day of each month in the year; to how much did her portion amount?

Here, before finding the amount of the series, we must find the last term, as directed in the rule after ex. 1.

Ans. $265 720 The two processes of finding the last term, and the amount, may, however, be conveniently reduced to one, thus:

When the first term, the ratio, and the number of terms, are given, to find the sum or amount of the series,-Raise the ratio to a power whose index is equal to the number of terms, from which subtract 1; divide the remainder by the ratio, less 1, and the quotient, multiplied by the first term, will be the

answer.

Applying this rule to the last example, 312 531441, and 531441 1 X 1265720. Ans. $265720, as before.

3 .1

12. A man agrees to serve a farmer 40 years without any other reward than 1 kernel of corn for the first year, 10 for the second year, and so on, in 10 fold ratio, till the end of the time; what will be the amount of his wages, allowing 1000 kernels to a pint, and supposing he sells his corn at 50 cents per bushel ?

1040-1

10-1

-X1=

S 1,111,111,111,111,111,111,111,111, 111,111,111,111,111 kernels.

Ans. $8,680,555,555,555,555,555,555,555,555,555,555 $555.