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EXAMPLES.

1. Given the least terin 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series:

2
8

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2. If the greatest term be 70, the common difference 3, and the number of terms 21, what is the least term, and the sum of the series?

Ans. The least term is 10, and the sum is 840. 3. A debt can be discharged in a year, by paying 1 shilling the first week, 3 shillings the second, and so on, always 2 shillings more every week; what is the debt, and what will the last payment be? Ans. The last payment will be 51 3s, and the debt is 1351 4s.

1

PROBLEM IV. To find an Arithmetical Mean Proportional between Two Given

Terms. Add the two given extremes or terms together, and take half their sum for the arithmetical mean required.

EXAMPLE. To find an arithmetical mean between the two numbers 4 and 14.

Here
14
4

2) 18

Ans.

9 the mean required.

PROBLEM

PROBLEM V.

To find Two Arithmetical Means between Two Given Extreines.

SUBTRACT the less extreme from the greater, and divide the difference by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, gives the means.

EXAMPLE

To find two arithmetical means between 2 and s.
Here 8

2

3) 6

Then 2 + 2 = 4 the one meane

and 4 + 2 = 6 the other mean.

com. dif. 2

PROBLEM VI.

To find any Number of Arithmetical Means between Two Given,

Terms or Extremes.

SUBTRACT the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required.

EXAMPLE.

To find five arithmetical means between 2 and 14.
Here 14

2

6) 12

Then by adding this com. dif. continually,

the means are found 4, 6, 8, 10, 12.

cam. dif. 2

See more of Arithmetical progression in the Algebra.

I 2

GFOMETRICAL

GEOMETRICAL PROPORTION and PROGRESSION.

In Geometrical Progression the numbers or terms have all the same multiplier or divisor. The most useful part of Geometrical Proportion, is contained in the following theorems.

THEOREM 1. , When four quantities are in geometrical proportion, the product of the two extremes is equal to the product of the two means.

Thus, in the four 2, 4, 3, 6, it is 2 x 6 = 3 x 4 = 12.

And hence, if the product of the two means be divided by one of the extremes, the quotient will give the other extreme. So, of the above numbers, the product of the means 12 = ? = 6 the one extreme, and 12 = 6 =2 the other extreme; and this is the foundation and reason of the practice in the Rule of Three.

THEOREM 2. In any continued geometrical progression, the product of the two extremes is equal to the product of any two means that are equally distant from them, or equal to the square of the middle term when there is an uneven number of terms.

Thus, in the terms 2, 4, 8, it is 2 x 8 = 4 X 4 = 16.
And in the series 2, 4, 8, 16, 32, 64, 128,

it is 2 x 128 = 4 x 64 = 8 x 32 = 16 x 16 = 256.

THEOREM 3. The quotient of the extreme terms of a geometrical progression, is equal to the common ratio of the series raised to the power denoted by 1 less than the number of the terms. Consequently the greatest term is equal to the least term multiplied by the said quotient.

So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common ratio is 2, and one less than the number of terms is 9; then the quotient of the extremes is 1024 = 2 = 512, and 29 :512 also.

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THEOREM 4. The sum of all the terms, of

any geometrical progression, is found by adding the greatest term to the difference of the extremes divided by i less than the ratio.

So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,

1024 - 2 (whose ratio is 2), is 1024 + = 1024 + 1022 = 2046.

2-1 The foregoing, and several other properties of geometrical proportion, are demonstrated more at large in the Algebraic part of this work. A few examples may here be added of the theorems, just delivered, with some problems concerning mean proportionals.

EXAMPLES

1. The least of ten terms, in geometrical progression, being 1, and the ratio 2 ; what is the greatest term, and the sum of all the terms ?

Ans. The greatest term is 512, and the sum 1023. 2. What debt may be discharged in a year, or 12 months, by paying 17 the first month, 21 the second, 41 the third, and so on, each succeeding payment being double the last; and what will the last payment be?

Ans. The debt 40951, and the last payment 20481.

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PROBLEM I.

To find One Goometrical Mean Proportional between any True

Numbers.

MULTIPLY the two numbers together, and extract the square root of the product, which will give the mean proportional sought.

EXAMPLE

To find a geometrical mean between the two numbers 3 and 12.

12

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PROBLEM II.

To find Two Geometrical Mean Proportionals between any Two

Numbers.

Divide the greater number by the less, and extract the cube root of the quotient, which will give the common ratio of the terms. Then multiply the least given term by the ratio for the first mean, and this mean again by the ratio for the second mean: or, divide the greater of the two given terins by the ratio for the greater mean, and divide this again by the ratio for the less mean.

EXAMPLE

To find two geometrical means between 3 and 24. Here 3) 24 (8; its cube root 2 is the ratio. Then 3 x..2 6, and 6 X 2= 12, the two means. Or 24 = 2 = 12, and 12 = 2 = 6, the same. That is, the two means between 3 and 24, are 6 and 12.

PROBLEM III.

To find any Number of Geometrical Means between Two Numbers,

Divide the greater number by the less, and extract such root of the quotient whose index is 1 more than the number of means required ; that is, the 2d root for one mean, the 3d root for two means, the 4th root for three means, and so on; and that root will be the common ratio of all the terms. Then, with the ratio, multiply continually from the first term, or divide continually from the last or greatest term.

EXAMPLE.

To find four geometrical means between 3 and 96. Here 3) 96 (32; the 5th root of which is 2, the ratio. Then 3x2=6, & 6x2 = 12, & 12 x 2 = 24, & 24 X 2 = 48. Or 96 ; 2=48,& 48 -2=24, & 24 -- 2 =12, &122=6.

That is, 6,12, 24, 48, are the four means between 3 and 96.

OF

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