series last given, the ratio or common multiplier might be considered to be . In what follows, however, the ratio will be taken always greater than a unit, according to the definition already given.' The following are the most useful rules for the management of quantities of this kind. RULE I. The first term and the ratio being given, to find any other proposed term (1.) Raise the ratio to a power whose index is equal to the number of the terms which precede the required term. (2.) Then, if it be an increasing series, multiply the first term by the result before found; otherwise divide it by that result. Exam. 1. Required the 8th term of the series of continual proportionals, whose first term is 6, and ratio 2. Here, the 7th power of 2 is found to be 128; which being multiplied by the first term 6, the product is 768, the 8th term. The reason of this operation will be manifest if it be considered, that in finding the successive terms up to the 8th, the first term must be multiplied by 2, the product by 2, that product by 2, and so on, till the 8th term would be found after seven such multiplications: and it is evident, that the same result will be found by a single multiplication by the 7th power of 2. A similar illustration serves in case of a decreasing series. Exam. 2. Required the 20th term of the series, whose first term and ratio are each 1·06. Here, we are to multiply the 19th power of 106 by 1·06, or, which is the same, we are to involve 106 to the 20th power. This is found by involution, to be 3.207135. Exam. 3. Required the 6th term of the `decreasing series whose first term is 100, and ratio 13. The 5th power of 13, or 1·5, is 7·59375, and 100 being divided by this, the quotient is 13-16872428, the term required. Ex. 1. Given the first term of an increasing series 12, and its ratio 3; to find the 18th term. Answ. 1549681956. 2. The first term of a decreasing series is 1, and the ratio 1:07; required the 14th term. Answ. 4149644. 3. The first term of an increasing series is £1943, and the ratio 105; required the 31st termi. Answ. £839-75333. 4. Given the first term of a decreasing series = 500, and the ratio 104; to find the 14th term. Answ. 300 287. It is proper for the learner to know, that in a series of continual proportionals, the product of the extremes is equal to the product of any two terms equally distant from them; or to the second power of the middle term, if the number of terms be odd. The reason of this is evident, since the greater extreine exceeds the term next it in the same ratio in which the other extreme is less than the term next it. 5. Given the first term of an increasing serfes = 1, and the ratio 2; required the 36th term. Answ. 34359738368. 6. Given the first term of an increasing series 1, and the ratio 3; required the 36th term. Answ. 50031545098999707. RULE II. To find the sum of a series of continual proportionals; multiply the greater extreme by the ratio, and divide the difference between the product and the less extreme, by the difference between the ratio and a unit. When the series is a decreasing one, and the number of terms infinite, divide the product of the ratio and the greatest term by the difference between the ratio and a unit. Or, Divide the ratio by the difference between it and a unit, and multiply the quotient by the first term.* Exam. 4. Given the first term of an increasing series = 4, the ratio 3, and the number of terms 6; to find the sum of the series. Here, by rule I. we find the last term to be 972. Multiplying this by the ratio, we obtain 2916: and dividing 2912, the difference between this and the first term, by 2, the difference, between the ratio and a unit, we obtain 1456, the required sum. The reason of this operation is best shown by Algebra; it may be illustrated in the following manner however: let the terms of the series be placed as in the 4+12+36 +108+324+972 = sum margin; then 12+36+108+324+972+2916 = sum × 3 let each term be multiplied by the ratio, and let the products be removed each one place towards the right hand. If the upper line be then subtracted from the lower, there will remain 2912 sum X2; and consequently the sum is equal to 2912÷2 = 1456. Now 2916 is evidently the product of the ratio and the greater extreme, and 2912 is the difference between this and the less extreme; also the divisor 2 is the difference between the ratio and a unit: and a similar illustration may be given in any other case. In a decreasing infinite series, the last term is to be regarded as nothing; and hence the reason of the rule for its summation is manifest. * The following rules, which may be illustrated in the same manner as the rule given in the text, may also on some occasions be employed with advantage: I. To sum an increasing series: (1.) Raise the ratio to a power whose index is equal to the number of terms: (2.) Divide the difference between the result and a unit, by the difference between the ratio and a unit: (3.) Multiply the quotient by the first term. II. To sum a decreasing series: (1.) Raise the ratio to a power whose index is equal to the number of terms: (2.) From the power thus obtained, take a unit, and divide the remainder by the same power: (3.) Divide the quotient by the difference between the ratio and a unit, and multiply the result by the first term, Exam. 5. Required the sum of the series whose least term is 15, and ratio 1.04, the number of terms being 31. This exercise may be wrought in the same manner as the last; or perhaps more easily by the first rule in the note. Thus, by involution, we find the thirty-first power of 104 to be 3.37313. Then 3.37313-1= 2·37313, and 104-104; also 2·37313 0459-32825, the product of which by 45, the first ten, is 2669-77125, the required sum. Exam. 6. Required the value of the interminate decimal 1'8'. This is the same as +10 +100000+ &c., continued without limit, where the ratio is evidently 100. Multiplying the first term therefore by 100, and dividing the result by 100 - 1, we obtain for the sum of the series, or the value of the decimal, , or in its lowest terms, . Ex. 7. Required the sum of the series whose first and last terms are 100, and 13-16872428 respectively, and its ratio 1. Answ. 273-66255144. 8. Given the extremes 1 and 18-42015, and the ratio = 1.06; required the sum of the series. Answ. 308·755983. 9. Find the sum of the infinite series whose greatest term is 100, and ratio 104. Answ. 2600. 10. Given the first term of an increasing series = 6, the ratio 4, and the number of terms 8; to find the sum of the series. Answ. 131070. 11. Given the first term of a decreasing series = 1, the ratio =3, and the number of terms = 12; to find the sum of the series. Answ. 1, or 1 nearly. 12. Given the least term of a series 1, the ratio = 1, and the number of terms 16: required the sum of the series. Answ. 1311.68167114. 13. Given the least term, the ratio terms 14: required the sum of the series. 14. Given the greatest term =12, the ratio= of terms 12; to find the sum of the series. 4, and the number of RULE III. The extremes and the number of terms being given, to find the ratio: Divide the greater extreme by the less, and extract that root of the quotient whose index is one less than the number of terms. Exam. 7. Given the extremes of a series 3 and 192, and the number of terms 7: required the ratio.. Here, 1923 64, the 6th root of which is 2, the ratio of the series. In the use of this rule we must generally employ either logarithms, or the rule given in page 211. 1 and 10, and the number of Answ. 1.333521. L 218 RULE IV. To find any proposed number of mean proportionals between two given numbers: (1.) Take the two given numbers as extremes; take also the number of terms in the series two greater than the required number of means, and find the ratio by rule III: (2.) Then the product of the ratio and the less extreme will be one of the means; the product of this mean and the ratio will be another; and thus all the means may be found, whatever is their number. When only one mean is required, it is most easily found by extracting the second root of the product of the extremes. Exam. 8. Find three mean proportionals between 5 and 1280. Here, the series would consist of 5 terms, and the extremes are 5 and 1280; and hence the ratio is found, by the last rule, to be 4; and by the repeated multiplication of this and of the first term, the means are found to be 20, 80, and 320. Exam. 9. Find a mean proportional between 5 and 10. Here, 5 × 10 = 50, the square root of which is 7·0710678, &c. the mean required. Ex. 16. Find two mean proportionals between 1 and 2. Answ. 1.259921, and 1.587401. 17. Find a mean proportional between 3.16227766. and 100. Answ. Ex. 18. If a thrasher agree to work 18 days for a farmer, on condition of receiving two grains of wheat for the first day's work, 6 grains for the second, 18 for the third, &c.: what would be the value of all he would be entitled to receive, supposing 7680 grains to fill a pint, and the wheat to be worth 7 shillings a bushel ? Answ. £275 17 5 19. Suppose a house, having 20 windows, to be sold at the rate of 4/0 for the first of these windows, 6/0 for the second, 9/0 for the third, and so on; the value of each being increased by one half of itself, to find the value of the next; for how much would it be sold? Answ. £1329 14 0§. 20. If a father give as a portion to his daughter, aged 19, a farthing for the first year of her life, three farthings for the second, 9 farthings for the third, &c.;. to how much does her portion amount? Answ. £605,344 10 34. 21. It is said that an Indian discovered the game of chess, and showed it to his sovereign, who was so much pleased with it, that The inven he desired the inventor to ask any reward he chose. tor then asked one grain of wheat for the first square of the chess table, two for the second, four for the third, and so on; doubling continually to 64, the number of squares. Now, suppose it had been possible for the prince to pay this reward, what would have been the value of the whole at 12/6 cwt., 10,000 grains being supposed to weigh a pound avoirdupois? Answ. £10,293,942,005,418 5 61. 22. Find the value of the interminate decimal 46′3. Answ. 17. 23. Required the value of the infinite decimal 51′85'. Answ. §‡. 24. Find the sum of the infinite series,,,, &c. Answ. 1. 25. Find the sum of the infinite series,,,, &c. Answ. §. HARMONICAL PROPORTION. It may be proper to subjoin to what has been said respecting equidif. ferent quantities and continual proportionals, a few observations on harmonical proportion, a subject which, though of minor importance, should not be entirely overlooked. Three or four numbers are said to be in HARMONICAL PROPORTION, when the first is to the last, as the difference of the first and second is to the difference of the last and the last but one. Thus, 2, 3, and 6, are three numbers in harmonical proportion, 2 being to 6 as 3-2 to 6-8: and 15, 12, 6, and 5, are four numbers in harmonical proportion, since 15: 5:: 15—12: 6—5. The reciprocals of three equidifferent numbers are in harmonical proportion. Thus,,, and the reciprocals of 2, 5, and 8, are in harmonical proportion: and if these fractions be reduced to equivalent ones having a common denominator, the numerators 20, 8, and 5, will be of the same nature. To find four numbers in harmonical proportion, find three such numbers in the manner just shown; and then let the mean and one extreme, and the mean and the other extreme, be multiplied or divided by any numbers whatever, and the four numbers thus found will be the terms required. Thus, taking 20, 8, and 5, the numbers last found, we have, by halving the first term and the mean, and trebling the mean and the third term, 10, 4, 24, and 15, which are in harmonic proportion, since 10: 15:: 10—4: 24-15; and in this manner we may find as many such numbers as we please. It is worthy of remark, that the reciprocals of continual proportionals are also continual proportionals, while the reciprocals of three equidifferent numbers are in harmonical proportion, and, consequently, the reciprocals of three numbers in harmonical proportion are equidifferent numbers. Farther also, if any two numbers be taken as extremes, the harmonical mean, the mean proportional, and the equidifferent mean between them are three continual proportionals. Thus, the three means between 20 and 5, are 8, 10, and 124: the second of which is a mean proportional between the other two, |