ART. 300. The two equations, l=ar"-1, and S= nish this general problem: Knowing any three of the five quantities a, r, n, l, and S, of a geometrical progression, to determine the other two. The following table contains all the values of each unknown quantity, or the equations from which it may be derived. REMARK. To determine the value of the unknown quantity in Nos. 3, 12, 14, and 16, may require the solution of an equation higher than the second degree. The values of n in the last four Nos. are obtained from the solution of an exponential equation (see Art. 382). Although the method of solving these equations has not been given, it was deemed proper to complete the table for the convenience of referThe pupil should be required to verify all the values except those here referred to. ence. EXAMPLES FOR PRACTICE. 1. Find the 8th term of the series 5, 10, 20, &c. Ans. 640. 2. Find the 7th term of the series 54, 27, 131, &c. Ans. Ans. 1. 3. Find the 6th term of the series 33, 21, 11, &c. 6. Of 1+3+9+, &c., to 9 terms. 7. Of 1+4+16+, &c., to 8 terms. 8. Of 8+20+50+, &c., to 7 terms. 9. Of 5+20+80+, &c., to 8 terms. 10. Of 1+3+9+, &c., to n terms. 11. Of 1-2+4-8+, &c., to n terms. 12. Of x−y++, &c., to n terms. Ans. 9841. Ans. 21845. Ans. 32497. Ans. 109225. Ans. (3-1). Ans. (12"). x+y . 13. The first term is 4, the last term 12500, and the numbe of terms 6. Required the ratio and the sum of all the terms. Ans. Ratio =5; Sum =15624. 4. Find the geometrical progression, when the sum of the first and second terms is 9, and the sum of the first and third 15. Ans. 3+6+12+, &c., 13—41+11⁄2—, &c. Find the sum of an infinite number of terms of each of the following series: 15. Of +++, &c. Ans.. 22. The sum of an infinite geometric series is 3, and the sum of its first two terms is 23; find the series. Ans. 8. 23. Find a geometric mean between 4 and 16. Let a=4, c=16, and m the required mean; then am::m:c; whence mac. 24. The first term of a geometric series is 3, the last term 96, and the number of terms 6; find the ratio, and the intermediate terms. By formula 13, page 251, we find_r=”-1 √ which in this case α becomes r=5/32-2; hence, the intermediate terms are 6, 12, 24, 48. If it be required to insert m geometrical means between two numbers a and b, we have n, the whole number of terms, hence, n-1=m+1, and r=m+1 25. Insert two geometric means between 19, and 2. m+2; Ans. 1. 26. Insert seven geometric means between 2 and 13122. Ans. 6, 18, 54, 162, 486, 1458, 4374. ART. 301. To find the value of Circulating Decimals, that is, decimals in which one or more figures are continually repeated. Circulating decimals are quantities in geometrical progression, where the common ratio is 10, 100, 1000, &c., according as one, two, or three figures recur; thus the circulating decimal .253131. . . . is equal to 31 31 31 is equal to 256+(+ and the part within the bracket is a geometrieal series, of which This operation may be performed more simply, as follows: Let S .253131. . . . Multiply by 100, in order to remove the decimal point to the commencement of the first period of decimals, we have 100S-25.3131. Again, multiplying by 100 to remove the decimal point to the commencement of the second period of decimals, we have Subtracting the preceding equation from the last, we get ART. 302. Three or more quantities are said to be in Harmonical Progression, when their reciprocals are in arithmetical progression. Thus, 1,,,, &c.; and 1,,,, &c. are in harmonical progression, because their reciprocals 1, 3, 5, 7, &c.; and 4, 31, 3, 21, &c. are in arithmetical progression. ART. 303. PROPOSITION.—If three quantities are in harmonical progression, the first term is to the third, as the difference of the first and second, is to the difference of the second and third. For if a, b, c, are in harmonical progression, arithmetical progression, 1 1 1 1 ... Hence, multiplying by abc, b α с ac-bc=ab—ac; or c(a—b)=a(b—c). Dividing both sides by a-b, and by a, we have Therefore, a Harmonical Progression is a series of quantities in harmonical proportion (Art. 280); or such that if any three consecutive terms be taken, the first is to the third, as the difference of the first and second is to the difference of the second and third. From this proposition it follows, that all problems with respect, to numbers in harmonical progression, may be solved by inverting them, and considering the reciprocals as quantities in arithmetical progression. This renders it unnecessary to give any special rules for the solution of problems in harmonical progression. EXAMPLES FOR PRACTICE. 1. Given the first two terms of a harmonical progression, a and b, to find the nth term. 1 1 Let 7 be the nth term, then (Art. 302), and are the first b α two terms of an arithmetical progression, and it is required to By means of this formula, when any two successive terms of a harmonical progression are given, any other term may be found. 2. Insert m harmonic means between a and b. Here, if d be the common difference of the reciprocals of the terms, we have ==2+(n−1)d, and d= a a-b a-b (n-1)ab (m+1)ab whence the arithmetical progression is found; and by inverting its terms, the harmonicals are also found. |