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2. A person bought a farm for £510 to be paid monthly in arithmetical progression, and to be completed in a year, each payment to exceed that preceding by £5: What were the first and last payments?

Ans. £15 the first payment, and £70 the last payment.

The following Table contains a summary of the whole doctrine of Arithmetical Progression.

Note. The table contains several cases, whose rules are not given in the text, because they are not very easily demonstrated without the aid of Algebra. Each of these cases however is illustrated by examples, which follow the table, so that the expression for the process in the table, may be more intelligible to the learner. It should be observed that where two letters or a figure with a letter or letters, occur in the rules in the table, without a sign between them, the product of the quantities is intended. Thus, d5 means d×5, and 8ds signifies 8XdXs.

CASES OF ARITHMETICAL PROGRESSION. Case Given | Required [

Solution

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EXAMPLES IN ARITHMETICAL PROGRESSION.

1. Given the first term 3, the common difference 2, and the sum of the series 399, to find the number of terms and the last

term.

By the 4th Case in the table we have,

√3X2--2)2+399×2×8—3×2−−2 ̧

2× 2

19, the number of terms.

And, √3x2-22+399×2×8-2-39, the last term.

2

2. Given the first term 3, the number of terms 19, and the sum of the terms 399, to find the common difference.

By the first rule of Case 6th, we have,

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3. Given the common difference 2, the, last term 39, and the sum of the series 399, to find the first term and the number of terms.

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4. A merchant owed to several persons $1080; to the greatest creditor he paid $142, to the greatest but one $132, and so on, in Arithmetical Progression; What was the number of creditors, and what did the least creditor receive?

Ans. The number of creditors was 15, and the least creditor received $2.

5. Given the last term 39, the number of terms 19, and the sum of the series 399, to find the common difference.

By the 2d rule in Case 8, we have,

2X19X39--399

19-1X19

=2, the common difference.

6. Sixteen persons gave in charity to a poor man in such a manner as to form an arithmetical series; the last gave 65 cents, and the whole sum was $5 60c.; What did each give less than the othfrom the last down to the first,

er,

Ans. 4 cents.

GEOMETRICAL PROPORTION.

THEOREM I.

IF four quantities, 2. 6. 4. 12, be in Geometrical Proportion, the product of the two means, 6×4 will be equal to that of the two extremes, 2×12, whether they are continued, or discontinued,* and, if three quantities, 2. 4. 8, the square of the mean is equal to the product of the two extremes.

THEOREM 2.

If four quantities, 2. 6. 4. 12, are such, that the product of two of them, 2×12 is equal to the product of the other two, 6X4, then are those quantities proportional.†

It was statel un ler Proportion in General, that the geometrical ratio of two quantities is expressed by the quotient, arising from dividing the antecedent by the consequent; thus, the geometrical ratio of 6 to 2 is 3,=g, and of 2 to 6, is

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or and of 3 to 8 is: and that in a proportion there must be two, or more, couplets which have equal ratios. Hence, four numbers will be geometrical proportionals, when the ratios, obtained in this manner, are equal. Thus 2, 4, 8, 16, are geometrically proportional, because 23 each to; and thus also, 9, 3, 12, 4, because each to 3. From these principles, it is easy to prove in a given example, the theorems in Geometrical Proportion. The factors should be kept separate by the sign of multiplication.

+

Let 2,4,3,6,be the geométrical proportionals; then 23. Multiply both fractions by the product of the second and fourth terms, and the fractions will obvi2X4X6 3X4X6 ously still be equal, and we haveThen cancel the equal

4

terms in the fractions, and 2×63x4, that is, the product of the extremes, 2X6, is equal to the product of the means, 4X3. The same may be shown in any other case, and, hence the general rule be inferred.

Again; Three numbers are geometrical proportionals, when the ratio of the first and second terms is equal to the ratio of the second and third. Thus, 2, 4, 8, are three geometrical proportionals, for each to. Proceed as 2×4×8 4X4X8 before, and we haveand 2×8=4×4, or 42, that is, the 4 product of the extremes, 2×8, is equal to the square of the mean, 4X4 or 43.

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+ Let the four quantities be 2, 6, 4, and 12, so that 2X12-6×4. Divide these equal products by the quantity 6×12, and the quotients will obviously 2×12 6X4

be equal, or

6X12 6×12"

Cancel the equal terms in these two fractions, and we have 3; whenee 2:6::4: 12, by the definition of geometrical pro portionals.

4

In the same way it may be shown, that if 2×8-4x4 or 43, then 2:4:: 48, and 2, 4, and 8, are three geometrical proportionals.

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THEOREM 3.

If four quantities, 2. 6. 4. 12, are proportional, the product of the means, divided by either extreme, will give the other extreme.* THEOREM 4.

The products of the corresponding terms of two Geometrical Proportions are also proportional.

That is, if 2: 6 :: 4 : 12, and 2: 4: : 5:10, then will 2×2: 6x4 4x5: 12 × 10 †

THEOREM 5.

If four quantities, 2, 6, 4, 12, are directly proportional

(1. Directly,

2. Inversely,

2:6 ::
6 : 2 ::

4 : 12 12:

4

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Because the product of the means, in each case, is equal to that of the extremes, and therefore the quantities are proportional by Theorem 1.

THEOREM 6.

If three numbers, 2, 4, 8, be in continued proportion, the square of the first will be to that of the second, as the first number to the. third; that is, 2×2: 4×4 :: 2 : 8.1

Let the four proportionals be 2, 4, 5, and 10; then 2×10=4×5, by Thec rem 1. Divide both expressions by 2, and

--

4X5

2X10

4X5

; or 10; or, di

2

2

vide, as before, by 10, and

2X10 4x5
10 10

-, or 2

4X5
10

that is, the product of the

means divided by one extreme, gives the other extreme. Hence, if the two means and one extreme be given, the other extreme, or geometrical proportional may be found.

+ Let there be given, 2: 6:4; 12, whence by Theorem 1: and also, 3:56:10, whence . Multiply the corresponding parts of these equal 2X3 6X4 10×12 Hence by the definition of geometrical propor

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tionals, 2x3: 6×5 :: 6×4 10X 12, and is the theorem.

Hence, if four quantities are proportional, their squares, cubes, &c. will likewise be proportional. Thus, let the terms be 2 : 6 :: 4 : 12, then 2×2:6×6 :: 4×4 : 12X12, or 22: 62: 42: 122, and hence also, 23:63: 43: 123 and 25: 65: 45: 125.

For since 2 4 4 8, thence will 2x8=4x4, by Theorem 1; and therefore 2×2×8=2x4x4, by equal multiplication; consequently, 2×2:4×4 :: 2 : 8, by Thecrem 2.

In like manner it may be proved that, of four quantities continually proportional the cube of the first is to that of the second, as the first quantity to the fourth.

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