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There are five particulars which belong to questions in geometrical progression; namely, 1. The least term, 2. The greatest term, 3. The number of terms. 4. The ratio. 5. The sum of all the terms. Any three of these five being given, the remaining two may be found, as follows.

} called the ertremes.

296. The least term, the greatest term, and the ratio being given, to find the sum of the series. RULE I. Multiply the greatest term by the ratio, and from the product subtract the least term for a dividend. 2. Subtract 1 from the ratio for a divisor. 3. Divide the dividend by the divisor, and the quotient will be the sum required.

ExAMPLEs.

1. The least term is 2, the greatest 6250, and the ratio 5, in a series in geometrical progression; required the sum of the series 2 Thus 6250 x 5 – 2 = 31250 – 2 = 31248 the dividend, And 5 – 1 = 4 the divisor.

Therefore 3.1248

2. The least term 1024, the greatest 59049, and the ratio 14, being given in a geometrical series, to find the sum of all the terms ? Thus 59049 x 14 – 1024 = 87.549.5 dividend. And 14 – 1 = y =.5 divisor. 87540.5 .Then to5 = 175099 the sum required. 3. Given in a geometrical progression the least term 10, the greatest term 10000, and the ratio 6, required the sum ?

= 7812 the sum required.

braic part of the work; the grounds of the several rules in arithmetical and geometrical progression, cannot be explained in a satisfactory manner by common arithmetic.

loooo x 6 – 10 59990 H--4. The least term is 1, the greatest 2187, and the ratio 3, required the sum of the series 2 Ans. 3280. 5. The extremes are 10 and 100, and the ratio 13, in a geometrical progression; required the sum ? Ans. 550.

Thus = 1 1998 the sum.

6. The extremes of a geometrical progression are 5 and 320, and the ratio 2; required the sum ? Ans. 635.

297. Given the greatest term, number of terms, and ratio, to find the least term. RULE. Involve the ratio to the power whose index is 1 less than the number of terms, divide the greatest term by this power, and the quotient will be the least term. 7. In a geometrical series, the greatest term is 972, the number of terms 6, and the ratio 3, required the least term 2

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8. Required the least term of a geometrical progression, of which the greatest term is 1536, the ratio 2, and the number

of terms lo 2 -
- 1536
Thus 10 – 1 = 9, then 29 = 512. wherefore

- the 512 3,

least term.

9. In a geometrical progression the greatest term is 10.2487, the ratio 1.1, and the number of terms 5; required the least term 2

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Ligi = 7, the least term. 10. In a geometrical progression consisting of 6 terms, the greatest term is 1024, and the ratio 4; required the least term 2 Ans. 1. - 11. Given the greatest term 768, number of terms 9, and ratio 2, to find the least term 2 Ans. 3.

298. The least term, ratio, and number of terms being given, to find the greatest term. RULE. Involve the ratio to that power whose index is one less than the number of terms; multiply the power by the least term, and the product will be the greatest term. 12. The least term of a geometrical progression is 3, the ratio 2, and the number of terms 9; to find the greatest term 2 Thus 9 – 1 = 8, then 2 * = 256, whence 256 x 3 = 768, the greatest term. 13. In a geometrical series of 5 terms, the least term is 10, and the ratio 7, required the greatest term 2 Thus 5 – 1 = 4, and 74 = 2401, whence 2401 x 10 = 24010, the greatest term. 14. The least term of a geometrical series is 8, the ratio 3, and the number of terms 7; to find the greatest term 2 Thus 3" x 8 = 729 × 8 = 5832, the greatest term. 15. In a geometrical progression there are given the least term 2, the ratio 3, and the number of terms 4; to find the greatest term 2 Ans. 54. 16. Required the greatest term of a geometrical series, whose least term is 5, ratio 6, and number of terms 7 2 Ans. 233280.

299. The two extremes, and the sum of the series being given, to find the ratio. RULE. Subtract the least term from the sum, and also the greatest from the sum; then divide the former remainder by the latter, and the quotient will be the ratio required. 17. In a geometrical progression, the extremes are 10 and 10000, and the sum is 11998, required the ratio 2 Thus 1 1998 – 10 = 11988. and 11998 – 10000 = 1998. 1 1988 1998 18. The extremes are 1024 and 59049, and the sum 175099; required the ratio 2 Thus 1750.99 — 1024 = 174075, and 175099 – 59049 = 116050, whence 174075 116050 19. To find the ratio of a geometrical progression, whose sum is 550, and the extremes 10 and 100 }

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Then

= 6, the ratio required.

= 13, the ratio required.

13, the ratio required.

20. The extremes are 1 and 2187, and the sum of the series 3280; required the ratio 2 Ans. 3.

21. The sum of a geometrical series is 635, and the extremes are 5 and 320, required the ratio 2 Ans. 2.

300. The least term and ratio being given, to find any proposed term of the series. Rule I. Write down a few of the leading terms of the given geometrical series, and place over them as indices the terms of an increasing arithmetical series, whose common difference is 1, namely 1, 2, 3, 4, 5, &c. when the least term and ratio of the given series are equal; and 0, 1, 2, 3, 4, 5, &c. when they are unequal. II. Add together such of the indices as will make the index of the term required; if the least term and ratio are equal, this index will be equal to the number denoting the place of that term; but if they are unequal, the index will be 1 less. III. Multiply those terms of the geometrical series together, which stand under the indices added, and the product will be the term sought, when the first (or least) term and ratio are equal '. IV. But if the first term be not equal to the ratio, involve the first term to the power whose index is 1 less than the number of terms multiplied, divide the above product by this power, and the quotient will be the term required.

22. The first term of a geometrical series is 2, the ratio 2, and the number of terms 14; required the last or greatest term 2 - OPERATION. Thus 1. 2. 3. 4. 5 indices. And 2. 4. 8. 16. 32 leading terms. Then 2 + 3 + 4 + 2 + 3 = 14 = inder of the 14th term.

* This property of the indices is the foundation of Logarithms; its use in this place is extremely obvious: for knowing the last term, we also know what its index will be; and knowing the index, we readily perceive what terms of the arithmetical series must be added together to produce it, and these terms indicate what terms in the geometrical series are to be multiplied together to produce the last term.

To try to account for this mutual correspondence of the two progressions, would be a vain attempt; like many other properties, it follows from the nature of numbers, and this is perhaps all that can be said on the subject.

VOL. I. T

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The first term and ratio being equal, I take the series 1. 2. 3. 4, &c. (beginning with 1) for indices: under these I place the leading terms 2.4. 8, &c. of the given geometrical series; then, because the index of the term required is evidently 14, I choose any of the indices, which added together make 14, namely, 2.3.4.2 and 3; I then multiply the terms which stand under these together, namely, 4.8. 16.4 and 8, and the product is the answer required.

23. Required the 20th term of the series 1.3. 9, 27, &c. :

OPERATION. Thus O. l. 2. 3. 4. 5 in dices. And 1. 3. 9. 27. 81. 243 leading terms. Then 5 + 5 + 4 + 3 + 2 = 19 inder of the 20th term. And 243 x 243 x 81 × 27 x 9 = 1162.358667 the 20th term.

Erplanation.

The first term and ratio not being equal, the indices must begin with 0, and consequently 19 will be the index of the 20th term. But by the rule, the first term ought to have been involved to the 4th power, (one less than 5, the number of terms multiplied,) by which the product of the terms should have been divided; this is omitted, because the first term being 1, all its powers will be 1, and dividing by l makes no alteration.

24. What is the 19th term of the series 5. 10. 20. 40, &c. 2

OPERATIon.
Thus O. l. 2. 3. 4 indices.
And 5. 10, 20.40, 80 leading terms.
Then 4 + 3 + 2 = 9 inder of the 10th term.
Whence 80 x 40 x 20 = 64000 dividend.

4 Also 5]* = 25 the divisor, wherefore 64000 = 2560 the 10th

term required.

o Erplanation. The first term 5, and ratio 2, being unequal, I divide the product of the

terms, viz. 64000 by 5), or 25: that is, by that power of the first term 5, whose index 2, is less by 1 than the number of terms 3, multiplied together.

25. What is the llth term of the series 1. 2. 4.8. 16, &c. 2 Arts. 1024.

26. Required the 13th term of the series 2.4, 8.16. 32, &c. 2 Afts, S192.

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