80-50-30 error too little. 80--65=15 error too little.

2. D, E and F, would divide 100L. among them, so as that E may have 3L. more than D, and F 4L. more than E; what is the share of each?

Ans. D's share 30L. E's 33L. F's 37L.

3. A, B and C, owe 1000L. of which B is to pay 100L. more than A, and C is to pay as much as both A and B: how much is each man's share of the debt?

Ans. A's share is 200L. B's 300L. and C's 500L. 4. Bought linen at 4s. per yard, and muslin at 2s. per yard; the number of yards of both was 8, and the whole cost 20s.: how many yards were there of each?

Ans. 2 yards of linen, and 6 yards of muslin. 5. The head of a certain fish is 9 inches long; its tail is as long as its head and half of its body; and the length of its body is equal to the length of its head and tail: what is the whole length? Ans. 6 feet.

6. A labourer hired for 40 days upon this condition, that he should receive 20 cents for every day he wrought, and should forfeit 10 cents for every day he was idle; at settlement he received 5 dollars. How many days did he work, and how many days was he idle? Ans. Wrought 30 days, idle 10.

7. A father dying, left to his three sons A, B, and C, his estate in money, dividing it as follows, viz. to A he gave half the estate, wanting 44L.; to B he gave a third of it, and 14L. over; and to C he gave the remainder, which was 821.. less than the share of B. What was the whole sum left, and what was each son's share? The sum left was 588L. of which A Ans. had 250L. B 210L. and C 128L.

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8. Two persons, A and B, have both the same income; A saves one fifth of his every year; but B, by spending 150 dollars per annum more than A, at the end of 8 years finds himself 400 dollars in debt: What is their income, and what does each spend per annum? Ans. Their income is 500 dollars per annum. A spends 400dols. and B 550.

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ARITHMETICAL PROGRESSION.

Any rank or series of numbers, increasing or decreasing by a common difference, is said to be in arithmetical progression; as 2, 4, 6, 8, 10, and 6, 5, 4, 3, 2, 1.

The numbers which form the series are called the terms. The first and last terms are called the extremes.

Note.-In any series of numbers in Arithmetical progression, the sum of the two extremes is equal to the sum of any two terms equally distant from them; as in the latter of the above series 6+1=4+3, and =5+2.

When the number of terms is odd, the double of the middle term is equal to the sum of the two extremes, or any two terms equally distant from the middle term; as in the former of the foregoing series 6×2=2+10, and =4+8.

CASE 1.

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

RULE.

1. Multiply the number of terms, less 1, by the common difference, and to the product add the first term, the sum is the last term.

2. Multiply the sum of the two extremes by the number of terms, and half the product will be the sum of all the terms.

EXAMPLES.

1. The first term of a certain series in arithmetical progression is 2, the common difference is 2, and the

number of terms 15; what is the last term, and the sum of all the terms?

2. Bought 15 yards of linen, at 2 cents for the first yard, 4 for the second, 6 for the third, &c. increasing 2 cents every yard; what was the cost of the last yard, and what was the cost of the whole?

Ans. The last yard cost 30cts. the whole cost $2.40. 3. Sold 20 yards of silk, at 3d. for the first yard, 6d. for the second, 9d. for the third, &c. increasing 3d. every yard; what sum did it amount to?

Ans. 2L. 12s. 6d.

4. Sixteen persons gave charity to a poor man; the first gave 5d. the second 9d. and so on in an arithmetical progression; how much did the last person give; and what sum did the man receive?

Ans. The last gave 5s. 5d. sum received 2L. 6s. 8d. 5. If 100 stones be laid two yards distant from each other in a right line, and a basket placed two yards from the first stone; what distance must a person travel to gather them singly into the basket?

Ans. 11 miles, 3fur. 180yds.

6. A merchant sold 1000 yards of linen, at 2 pins for the first yard, 4 for the second, and 6 for the third, &c. increasing two pins every yard; how much did the linen produce, when the pir pins were afterwards sold at 12 for a farthing?

CASE 2.

Ans. 86L. 17s. 10d.

When the two extremes and number of terms are given, to find the common difference.

RULE.

Divide the difference of the extremes by the number of terms, less one; the quotient will be the common difference.

EXAMPLES.

1. Twenty and sixty are the two extremes of a certain series in arithmetical progression, and 21 is the number of terms; what is the common difference?

Ans. 2.

60

20

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extremes.

21-1-20)40 Difference of extremes.

1

2 Common difference.

2. There are 21 men whose ages are equally distant from each other in arithmetical progression; the youngest is 20 years old, and the eldest 60; what is the common difference of their ages, and the age of each man? Ans. Common difference 2 years.

60 years is the age of the first man.

60-2-58

58-2-56

age of the second. age of the third, &c.

3. A debt is to be paid at 16 different payments in arithmetical progression; the first payment to be 14L. and the last 100L.: what is the common difference, each payment, and the whole debt?

Ans.

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Common difference 5L. 14s. 8d. First payment 14L. Second, 19L. 14s. Sd. Third, 25L. 9s. 4d. &c. 4. A person is to travel from Philadelphia to a certain place in 16 days, and to go but 4 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 79 miles; what is the common difference; and what the whole distance?

Common difference 5 miles.
Distance 664 miles.

GEOMETRICAL PROGRESSIOΝ.

Any series of numbers, the terms of which increase by a common multiplier, or decrease by a common divisor, are said to be in geometrical progression; as 3, 6, 12, 24, 48; and 48, 24, 12, 6, 3.

The multiplier or divisor by which the series is increased or decreased is called the ratio. The last term and sum of the series are found by this

RULE.

1. Raise the ratio to the power whose index is one less than the number of terms given, which, being multiplied by the first term, will give the last term, or greater extreme.

2. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by ratio less one for the sum of the series.

EXAMPLES.

1. A thresher wrought 20 days, and received for the first day's labour 4 grains of wheat; for the second, 12; for the third, 36, &c. How much did his wages amount to, allowing 7680 grains to make a pint, and the whole to be disposed of at one dollar per bushel?

Note. The first term in this question is 4, the ratio 3, the number of terms 20: therefore raise the ratio to the 19th power, which is one less than the number of