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GEOMETRICAL PROPORTION.

THEOREM 1.

IF four quantities, a. b. c. d. (2. 6. 4. 12.) be in Geometrical Proportion, the product of the two means, be (6x4) will be equal to that of the two extremes, ad (2×12) whether they are continaend discontinued, and, if three quantities, a. b. c. (2. 4. 3.) the square of the mean is equal to the product of the two extremes.

*

THEOREM 2.

or

If four quantities, a. b. c. d. (2. 6. 4. 12.) are such, that the product of two of them, ad, (2×12) is equal to the product of the other two, bc, (6X4) then are those quantities proportional.†

THEOREM 3.

If four quantities a. b. c. d. (2. 6. 4. 12.) are proportional, the rectangle of the means, divided by either extreme, will give the oth

er extreme.

THEOREM 4.

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

That is, if a: b::c: d ( 2 : 6 :: 4: 12,) and e: f::g: h (2: 4: 5: 10,) then will ae: bf:: cg: dh (2×2 : 6×4 :: 4×5: 12×10.)§ THEOREM 5.

For fince the ratio of a to b (2 to 6) or the part, which a is of b (2 is of 6) is ex

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fince, by fuppofition, the two ratios are equal, let them both be multiplied by bd,

a

2

C

4

(6×12) and the products—X64 (—×6×12) and—X6d÷x6×12) will likewife.

b

d

12

2X6X12 4X6X12

6

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6

--- =

12

-, or, 2×12=6X4.)

For fince, by fuppofition, the products ad (2×12) and be (6×4) are equal, let

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both be divided by bd (6×12) and the quotients — (—) and

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bd (b)

bd (d) 6×12 (6) 6×19

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For by the second Theorem, ad=bc(2×12=6×4) whence dividing both fides of

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the equation by a (2) we have d——(12— 12=-- Hence, if the two means and one

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that is, ae: bf:: eg: db (2X2: 6×4:: 4×5: 12×10:) Hence it follows, that if any quantities be proportional, their squares, cubes, &c. will likewise be propor▾ tional

THEOREM 5.

If four quantities, a. b. c. d. (2, 6. 4. 12.) are directly proportional,

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4:12)

a b c d (2 : 6 ::
bad: c (6 : 2 :: 12: 4)
a: cbd (2 : 4 ::

6:12)

a: a+b: cc+d(2 : 8 ::

4 : 16)

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(r

r

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

rem 1.

THEOREM 6.

If three numbers, a. b. c. (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, a2: b2 :: a: c (2X2 : 4X4 :: 2 : 8.)*

THEOREM 7.

In any continued Geometrical Proportion (1. 3.9 27. 81. &c.) the product of the two extremes, and that of every other two term equally distant from them are equal.†

THEOREM 8.

The sum of any number of quantities, in continued Geometrical Proportion, is equal to the difference of the rectangle of the second and last terms, and the square of the first, divided by the difference of the first and second terms.‡ GEOMETRICAL.

:

For fince a: b :: b c (2 : 4 :: 4: 8) thence will ac=bb (2×8=4×4) by Theorem 1; and therefore aac-abb (2×2×8=2×4×4) by equal multiplication; confequently, a2 b2 :: a: c (2X2: 4x4 2: 8) by Theorem 2.

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

For, the ratio of the first term to the fecond being the fame as that of the last but one to the last, these four terms are in proportion; and therefore by Theorem 1, the rectangle of the extremes is equal to that of their two adjacent terms; and after the fame manner, it will appear that the rectangle of the third and last but two is equal to that of their two adjacent terms, the fecond and last but one, and fo of the reft; whence the truth of the propofition is manifeft.

For, let the first term of the proportion be denoted by a, the common rako by r, the number of terms by n, and the fum of the whole feries by s, then it is plain that the second term will be expreffed by a×r, or, ar; the third by ar×1, of

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n-1

ar; the fourth by ar Xr, or, ar; and the nth, or aft term, by ar ; and there

2

3

fore the proportion will stand thus, a+aṛ+ar +ar +ar +ar

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equation multiplied by r, gives ar+ar tar tar .....tar

the first equation being subtracted, there will remain-a+ar =rs—s: Whence,

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GEOMETRICAL PROGRESSION.

A GEOMETRICAL Progression is, when a rank or series of numbers increases, or decreases, by the continual multiplication, or division, of some equal number.

PROBLEM I.

Given one of the extremes, the ratio, and the number of the terms of a geometrical series, to find the other extreme.

RULE.-Multiply, or divide, (as the case may require) the given extreme by such power of the ratio, whose exponent* is equal to the number of terms lefs 1, and the product or quotient, will be the oth

er extreme.

EXAMPLES.

ever, as 2. 6. 18. 54. 162. 486. and their fum will be 2+6+18+54+162+486 728: This equation multiplied by the ratio, will ftand thus, 6+18+54+162+486 +1458=2184: now it is plain that the fum of the fecond feries will be fo many times the first, as is expreffed by the ratio; subtract the first feries from the fecond, and it will give 1458—2—2184-728, which is evidently so many times the 1458-2 2916-4 fum of the first series, as is expressed by the ratio less 1; whence

as was to be demonstrated.)

8-1

6-2

As the laft term or any term near the laft, is very tedious to be found, by continual multiplication, it will often be neceffary in order to ascertain them, to have a series of numbers in Arithmetical Proportion, called indices, or exponents, beginning with a cypher, or an unit whose common difference is one.

When the firf term of the feries and the ratio are equal, the indices must begin with an unit, and, in this cafe, the product of any two terms is equal to that term, fignified by the fum of their indices.

1. 2. 3. 4. 5. 6, &c. indices, or arithmetical feries.

Thus, 2. 4. 8. 16. 32. 64, &c. geometrical feries (leading terms.)

Now, 6+6=12=the index of the twelfth term, and

64x64-4096=the twelfth term.

But, when the firft term of the feries and the ratio are different, the indices must begin with a cypher, and the fum of the indices, made choice of, must be one lefs than the number of terms, given in the question; becaufe 1 in the indices ftands over the fecond term, and 2 in the indices, over the third term, &c. And, in this case, the product of any two terms divided by the firft, is equal to that term beyond the first, fignified by the sum of their indices.

3. So. 1. 2. 4. 5. 6, &c. indices. Thus, 21. 3. 9. 27. 81. 243. 729, &c. geometrical series.

Here, 6+ 5 = 11 the index of the 12th term.

729×243=177147 the 12th term, because the first term of the feries and the ratio are different, by which mean a cypher stands over the first term. Thus, by the help of these indices, and a few of the first terms in any geometrical feries, any term, whofe distance from the first term is affigned, though it were ever so remote, may be obtained without producing all the terms.

Note. If the ratio of any geometrical series be double, the difference of the greatest and leaft terms is equal to the fum of all the terms, except the greatest; if the ratio be triple, the difference is double the fum of all but the greateft; if the ratio be quadruple, the difference is triple the fum of all but the greatest, &c.

In

any feries in decreafing to infinity-if the fquare of the firft term be divided by the difference between the first and fecond, the quotient will be the fum of the feries.

EXAMPLES.

1. If the first term be 4, the ratio 4, and the number of terms 9: What is the last term?

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4. 16. 64. 256.×256=65536=power of the ratio, whose exponent is less by 1, than the number of terms.

65536=8th power Multiply by 4=first term.

262144 last term.

Or, 4x4*=262144-the_ Answer.

of the ratio.

2. If the last term be 202144, the ratio 4, and the number of terms 9, what is the first term?

Last term.

8th power of the ratio 48-65536)262144(4=the first term.

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Again, given the first term, and the ratio, to find any other term assigned.

RULE I.

When the indices begin with an unit.

1. Write down a few of the leading terms of the series, and place their indices over them.

2. Add together such indices, whose sum shall make up the entire index to the term required.

3. Multiply the terms of the geometrical series, belonging to those indices, together, and the product will be the term sought.

EXAMPLES.

1. If the first term be 2, and the ratio 2, what is the 13th term? 1. 2. 3. 4. 5+5x3= 13

2. 4. 8. 16. 32x32x8=8192 \ns.

Or, 2×218192.

2. A merchant wanting to purchase a cargo of horses for the West-Indies, a jockey told him he would take all the trouble and expence upon himself, of collecting and purchasing 30 horses for the voyage, if he would give him what the last horse would come to by doubling the whole number by a half penny, that is, two farthings for the first, four for the second, eight for the third, &c to which, the merchant, thinking he had made a very good bargain, readily agreed: Pray what did the last horse come to, and what did the horses, one with another, cost the merchant?.

1. 2. 3. 4. 5. 6+6=12th.

12+12+ 6=last term.

2. 4. 8. 16. 32. 64x64-4096, and 4096×4096×64=

1073741824 qrs.=.1118481 1s. 4d. and their average price was .37282 14s. Od. a piece.

RULE II.

When the indices begin with a cypher.

1. Write down a few of the leading terms of the series, as before, and place their indices over them.

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