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meaning in this term, we choose to call the series, what it is in truth, a series of Continual Proportionals.

The common multiplier, or common divisor, by which the successive terms are increased or deminished, is called the ratio of the series, or the common ratio.

1 2 3 4 5 6

Thus, 3, 6, 12, 24, 48, 96 is a series of continual proportionals, in which each successive term is produced by multiplying the preceding term by 2, which is the common ratio. The numbers 1, 2, 3, 4, &c. standing above the series, mark the place, which each term holds in the series.

Also, 729, 243, 81, 27, 9, 3, 1, is a series of continual proportionals, in which each successive term is found by dividing the preceding term by 3, the common ratio.

In an increasing series, the ratio is the quotient, which results from the division of the consequent by the ante cedent; but in a decreasing series, the ratio is the quo tient resulting from the division of the antecedent by the consequent.

In every series of continual proportionals, any four successive terms constitute a proportion. Thus, in the first of the above series, 3: 6=12:24, and 6 ; 12—24 : 48, also, 12 24 48 96. In the second series, 729: 243=81; 27, 243: 81=27: 9, 81: 27=9: 3, 27: 9=3:1. Therefore, when there are only four terms, the product of the extremes is equal to the product of the means.

Furthermore, in any series of continual proportionals, the product of the extremes is equal to the product of any two terms equally distant from them; and equal to the second power of the middle term, when there is an uneven number of terms. For instance, take the continual proportionals 2, 4, 8, 16, 32, 64, 128; then 2 X 128 4X 64; also 2 X 128 8 X 32; and 2× 128= 16 X 16.

When the first term and the ratio are given, a series of continual proportionals may be extended to any number of terms by continually multiplying by the ratio in an increasing series, or dividing in a decreasing series,

For example, the first term being 2, and the ratio 3, if we make it an increasing series by continually multiplying by the ratio, we obtain the following series, 2, 6, 18, 54, 162, 486, which may be extended to any number* of terms; but, if we make it a decreasing series by continu ally dividing by the ratio, we obtain the following series, 2, 3, 3, 27, 81, 243, which may also be extended to any number of terms.

2

In the series 2, 6, 18, 54, 162, 486, we obtain the second term by multiplying the first term by the ratio; the third term by multiplying the second term by the ratio; the fourth term by multiplying the third term by the ratio; the fifth term by multiplying the fourth term by the ratio; the sixth term by multiplying the fifth term by the ratio. Therefore, since to obtain the sixth term, we have to multiply five times by the ratio, it is evident that we should also obtain the sixth term by multiplying the first term by the fifth power of the ratio. The fifth power of 3 is 243, which being multiplied into the first term, the result is 486, the same as in the series.

Hence we see, that any term in any increasing series of continual proportionals may be found by multiplying the first term by that power of the ratio, which is denoted by the number of terms preceding the required one. For instance, the ninth term in an increasing series is found by multiplying the first term by the eighth power of the ratio; thus let 2 be the first term, and 3 the common ratio; then 2X 38 gives the ninth term, which is 13122.

If the series be a decreasing one, any term in it may be found by dividing the first term by that power of the ratio, which is denoted by the number of terms preceding the required one. For instance, the seventh term in a decreasing series is found by dividing the first term by the sixth power of the ratio; thus, let 24576 be the first term in a decreasing series, and 4 the common ratio; then 24576÷4° gives the seventh term, which is 6.

We will now state several problems, which occur in continual proportionals, and give the rules for performing them.

PROBLEM I. The first term and the ratio being given, to find any other proposed term.

RULE. Raise the ratio to a power, whose index is equal to the number of terms preceding the required term: then, if it be an increasing series, multiply the first term by this power of the ratio; but, if it be a decreasing series, divide the first term by it: the result will be the required

term.

1. Required the eighth term in an increasing series whose first term is 6, and ratio 2.

2. Required the ninth term in a decreasing series, the first term of which is 131072, and the ratio 4.

3. What is the seventh term in an increasing series. the first term being 3, and the ratio 1.5?

4. What is the sixth term in an increasing series, whose first term is 16307, and ratio 7?

5. What is the tenth term in a decreasing series, the first term being 387420489, and the ratio 9?

One of the principal questions, which occurs in a series of continual proportionals, is to find the sum of the series. We shall, therefore, illustrate the method.

Let there be given the following series, consisting of seven terms, whose common ratio is 3; viz. 2, 6, 18, 54, 162,486, 1458. Let each term in this series be multiplied by the ratio 3; and let each product be removed one place to the right of the terms in the given series. The given series, 2, 6, 18, 54, 162, 486, 1458 multiplied by ratio. 6, 18, 54, 162, 486, 1458, 4374.

Now the last term in the second series is produced by multiplying the last term in the given series by the ratio; and it is evident that if the given series be subtracted from the second series, the remainder will be the last term in the second series diminished only by the first term in the given series, and this remainder will be twice the sum of the given series; consequently, if we divide 't by 2, the quotient will be the sum of the given series; but 2 is the ratio less 1. Hence

PROBLEM II. The extremes and the ratio being given, to find the sum of the series.

RULE. Multiply the greater extreme by the ratio, from the product subtract the less extreme, and divide the remainder by the ratio less 1, and the quotient will be the sum of the series.

6. The first term in a series of continual proportionals is 1, the last term is 65611, and the ratio is 3. What is the sum of the series?

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7. The extremes of a series of continual proportionals are 3 and 12288, and the ratio is 4.

of the series?

What is the sum

8. The first term in a series of continual proportionals in 12500, the last term is 4, and the ratio 5. What is the sum of the series?

9. The first term in a series of continual proportionals is 7, the last term 1792, and the ratio is 2.

sum of the series?

What is the

10. The extremes in a series of continual proportionals are 5 and 37.96875, and the ratio is 1.5. What is the sum of the series?

11. The first term in a series of continual proportionals is 100, the last term .01, and the ratio 2.5. Required the sum of the series.

PROBLEM III. The first term, the ratio, and the number of terms given, to find the sum of the series.

RULE. Find the last term by problem 1, and the sum of the series by problem 2.

12. The first term in an increasing series of continual proportionals is 6, the ratio 4, and the number of terms 8. What is the sum of the series?

13. The first term in an increasing series of continual proportionals is, the ratio 4, and the number of terms 13. Required the sum of the series,

14. The first term in a decreasing series of continual proportionals is 1, the ratio 3, and the number of terms 12. What is the sum of the series?

15. A man offers to sell his horse by the nails in his shoes, which are 32 in number. He demands one mill for the first nail, 2 for the second, 4 for the third, and so on, demanding for each nail twice the price of the preceding. It is required to find what would be the price of the horse.

16. An ignorant fop wanted to purchase an elegant house, and a facetious man told him he had one, which he would sell him on these moderate terms; viz. that he should give him one cent for the first door, 2 for the second, 4 for the third, and so on, doubling the price for every door, there being 36. It is a bargain, cried the simpleton, and here is a half-eagle to bind it. What was the price of the house?

PROBLEM IV. The extremes and the number of terms being given, to find the ratio.

RULE. Divide the greater extreme by the less, and the quotient will be that power of the ratio, which is denoted by the number of terms less 1; consequently, the corresponding root of this quotient will be the ratio.

This problem is the reverse of problem 1, and the reasoning which precedes that problem, sufficiently elucidates the rule in this.

The first term in a series of proportionals is 192, the last term 3, and the number of terms 7. What is the ratio? 192 3 = 64; the number of terms less 1, is 6; therefore the sixth root of 64, which is 2, is the ratio. 17. In a series of continual proportionals, the first term is 7, the last 45927, and the number of terms 9. What is the ratio?

18. The first term in a series of continual proportionals is 26244, the last term 4, and the number of terms 5. Required the ratio.

19. The first term in a series of continual proportionals is, the last term 1029427, and the number of terms 8. What is the ratio ?

20. The first term in a series of continual proportionals is 78125, the last term 15, and the number of terins 11. Required the ratio.

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