DUODECIMALS APPLIED TO MEASURING WOOD.

Q. How is wood estimated, in buying and selling?
A. By the cord, or by the cord-foot.

Q. What do you understand by a cord-foot of wood? A. Sixteen solid or cubic feet; that is, 4 feet long, 4 feet wide, and one foot high, make a cord-foot.

Q. How many such cord-feet make one cord of wood? A. Eight; that is, 8 feet long, 4 feet wide, and 4 feet high, make one cord.

Q. How do you measure a load of wood, in a cart or wagon? A. If the wood be of customary length, (8 feet,) multiply the average width by the height, and half the product will be the number of cord-feet.

Q. If the load be not 8 feet long, how do you measure it? A. Multiply the length by the average width, and that product by the height, then divide this last product by 16, and the quotient will be the number of cord-feet.

Q. How do you find the number of cords of wood, contained in any number of cord-feet?

A. Divide the number of cord-feet by 8, and the quotient will be the number of cords.

EXAMPLES.

1. How many feet of wood, in a load 3 feet, 9 inches wide, 4 feet, 3 inches high, and of full length?

2. How many feet, in a load of wood, 7 feet long, 4 feet, 3 inches wide, and 5 feet, 9 inches high; and what must you pay for it, at $5,50 cents a cord?

Operation. 7 ft. X4 ft. 3'x5 ft. 9'-16=11 ft. 5 in.+ Ans. 11 ft. 5 inches, and will cost $7,848 m. 3. How much wood in a load, and what must you give for it, measuring 3 ft. 4 in. in width, 2 ft. 6 in. high, and full length, paying at the rate of $6 the cord?

Ans. 4 ft. 2 in. Must give $3,12}.

4. How much wood in a load, 12 ft. long, 3 ft. 8 in. wide, and 4 ft. 10 in. high; and how much must I pay for it, at $6,50 cents a cord? Ans. 13 ft. 3 in. Must pay $10,799 m.

5. A poor woman bought a load of wood, measuring 3 ft. in length, 2 ft. 8 in. in width, and 24 feet in height; how much wood was there, and what must she pay for it, at the rate of $4,373 a cord? Ans. 1 ft. 4 in. Amount, 74 cents, 4 m.+ 6. How much wood can be put into a wagon, whose body is 8 ft. 9 in. long, 4 ft. 3 in. wide, and the stakes 5 ft. 8 in. high; and what will it be worth, at $63 a cord?

Ans. 1 cord, 5 ft. 2 in. Is worth $10,49 cts.+

7. Bought a load of wood, measuring 7 ft. 8 in. in length, 3 ft. 9 in. in width, and 5 ft. 7 in in height; how many feet of wood in the load, and what must I pay for it, at the rate of $7 a cord. Ans. 1 cord, 2 feet. Must pay for it $8,75+. 8. If I buy a load of wood, measuring 7 ft. 8 in. in length, 4 ft. 6 in. wide, and 3 ft. 10 in. high, how many feet does it contain, and what must I pay for it, at the rate of $6,75 cents the cord? Ans. 84 feet. Must pay $6,96.

9. Two persons bought a load of wood together; A.'s end of the load measured 3 ft. 9 in. wide, 4 ft. 6 in. high, and 4 ft. 3 in. long; B.'s end measured 3 ft. 7 in. wide, 4 ft. 11 in. high, and 4 ft. 3 in. long; how much wood had each, and what must each pay, at the rate of $5,50 a cord?

Ans.

A. had 4 feet, 5 inches+. Must pay $3,07+.
B. had 4 feet, 8 inches+. Must pay $3,20.+.

ARITHMETICAL PROGRESSION.

Q. What is an Arithmetical Progression?

A. It is any rank of numbers, more than two, increasing by a common excess, or decreasing by a common difference.

Q. What is such a rank of numbers, as 2:4:6:8:10: 12, &c. called?

A. It is called an ascending arithmetical series.

Q. What is such a rank of numbers, as 12: 10:8: 6:4:2 &c. called?

A. It is called a descending arithmetical series.

Q. What is to be understood, when you speak of the terms of an arithmetical progression?

A. It is the numbers, or rank of numbers, that form the series.

Q. How many parts are always included in an arithmetical

series ?

S*

A. Five parts, viz.: 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms.

Q. What is meant by the extremes of a progression ?

A. The first term and the last term are called the two extremes. The other terms are called the means.

Q. How many parts of an arithmetical series, must always be given, to find any other part required?

A. Three. By having any three parts given, the other two may be easily found.

CASE FIRST.

Q. What is the first case in arithmetical progression ? A. It is when one of the extremes, the common difference, and the number of terms of an arithmetical series, are given, to find the other extreme.

Q. What is the RULE in this case?

A. Subtract one from the number of terms, and multiply this remainder by the common difference; then (if the greatest or last term be required) add the first, or least term to the product, and the sum will be the last term, or answer; but, if the first or least term be required, subtract the product from the greatest or last term, and the remainder will be the first term.

EXAMPLES.

1. If the least term of an arithmetical series be 3, the common difference 2, and the number of terms 9, what will be the greatest or last term?

Operation. 9-1=8 number of terms less by 1.

x2= common difference.

16

+3 the first or least term.

Ans. 19 the greatest or last term.

2. If the greatest term of a series be 70, the common difference 3, and the number of terms 21, what is the least term? Operation. 21-1-20= number of terms, less by 1. X3 common difference.

Therefore

60

70-60-10 the first term, or answer.

3. If a man pay one dollar the first week, 3 dollars the second week, and continue to increase his payment 2 dollars, every week for the term of one year; what will be the amount of his last, or 52d payment?

Ans. $103.

ARITHMETICAL PROGRESSION.

223

4. A debt was discharged in a year, by paying a certain sum the first week, and increasing each weekly payment by $2, the last payment being $103; what was the first? Ans. $1.

5. A. owes to B. a certain sum of money, which he agrees to pay in arithmetical progression; the first payment to be $3, the number of payments 24, and the common difference 2; what will be the amount of the last payment? Ans. $49. 6. If a man travel 15 days, and increase each day's journey 4 miles, and the last day travel 75 miles, how far must he have travelled the first day? Ans. 19 miles.

CASE SECOND.

Q. What is the second case in arithmetical progression ? A. It is when the two extremes and common difference are given, to find the number of terms?

Q. What is the RULE in this case?

A. Divide the difference of the extremes, by the common difference, add 1 to the quotient, and the sum will be the number of terms.

EXAMPLES,

1. If the two extremes are 3 and 27, and the common difference 2, what is the number of terms?

Operation. 27-3÷2+1=13 the number of terms.

2. A man travelled 15 miles the first day, and increased his journey 3 miles a day, until he travelled 75 miles in a day; how many days did he travel? Ans. 21 days.

4. A man discharged a debt by weekly payments, and paying one dollar the first week, increasing his payment 5 dollars each week, until his last payment was 81 dollars; how many payments did he make ?

CASE THIRD.

Ans. 17.

Q. What is the third case in arithmetical progression? A. It is when the extremes and number of terms are given, to find the common difference.

Q. What is the RULE in this case?

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

EXAMPLES.

1. In an arithmetical series, the extremes are 3, and 27, and the number of terms, 13; what is the common difference?

Operation: 27-313-1-2, the common difference.

2. The two extremes of a series are 7, and 67, and the num. ber of terms 16; what is the common difference? Ans. 4. 3. A person on a journey, travelled 21 days, going 15 miles the first day, and 75 miles the last; what was his daily inAns. 3 miles.

crease?

4. A debt was paid in 17 weeks, by paying one dollar the first week, and 81 dollars the last; what was the weekly in. crease of the payments ? Ans. 5 dollars.

CASE FOURTH.

Q. What is the fourth case in arithmetical progression? A. It is when the two extremes, and the number of terms are given, to find the sum of the series.

Q. What is the RULE in this case?

A. Add the two extremes together, and multiply their sum by the number of terms, and half the product will be the answer, or sum of the series.

EXAMPLES.

1. The first term of an arithmetical series is 3, the last term 63, and the number of terms 21; what is the sum of the series? Operation. 63-+3-66. Then 66×21÷2-693 Ans. 2. A man travelled 15 days, in arithmetical progression, going 15 miles the first day, and 85 miles the last day; how many miles did he travel in all? Ans. 750 miles. 3. A merchant sold 26 yards of broadcloth, the first yard for 12 cents, the last for $7,873, arithmetical; what was the common difference, and what did the whole amount to?

Ans. Common difference, 31 cts. Whole amount, $104. 4. A man sold a farm of 100 acres, at $5 for the first acre, increasing the price of each succeeding acre, by $3; what did the last acre sell for, and to what did the whole amount?

Ans. The last acre sold for $302. Whole amount $15350. 5. A merchant sold 25 yards of cloth, in arithmetical progression, the first yard for 10 cents, the last yard for $2,50; what was the rate of increase, and to what did the whole amount? Ans. Common difference, 10 cts. Amount, $32,50.

6. A person bought 50 acres of land, at $75 an acre; a gentleman offered to purchase it, by paying $10 for the first acre, and increasing the price of each acre, by $5 arithmetical; what was the average price per acre, and would the man make, or lose, and how much, by selling his farm at that rate?

Ans.{

Average price per acre,
He would make $2875.

$132,50