Thus, the sum of the 4th powers sx sum of the cubes-px sum of the squares. The sum of the 5th powers=s× sum of the 4th powers—px sum of the cubes. The sum of the 6th powers=s× sum of the 5th powers—px sum of the 4th powers, &c. &c. Hence the sum of the nth powers of x and y will be as follows; n-2 n-3 n- -4 x"+y"=s" ―n.s p+n. .S p-n. +n. 2 .S p1-, &c. n-5 n-6 n-7 n-8 2 3 4 n-4n-5 n-6 13. To investigate the rules of arithmetical progression. Let a the least term z=the greatest } called also the extremes. n=the number of terms d=the common difference of the terms s=the sum of all the terms. Then will a+a+d+a+2d+a+3d+, &c. to a+n-1.d be an increasing series of terms in arithmetical progression. And z+z-d+z−2 d+z−3d+, &c. to z―n-1.d will be a decreasing series in arithmetical progression. 14. Now since in the increasing series a+n-1.d=the greatest term, and z=the greatest term by the notation, therefore z=a+ n-1.d (THEOREM 1.) Whence by transposition, &c. a=z-n-1.d 2-a z-a (THEOR. 2.) d= (THEOR. 3.) and n=— +1 (THEOR. 4.) n- 1 d Whence, of the first term, last term, number of terms, and difference, any three being given, the fourth may be found by one of these four theorems. 15. Next, in order to find s, and to introduce it into the foregoing theorems, let either of the above series, and the same series inverted be added together; and since the sum of each series is= s by the above notation, the sum of both added together, will evidently be 2s. Thus, The series a+a+d+a+2d+a+3d+&c.—s. The series inverted a+3d+a+2d+a+d+a. =S. Their sum ... 2a+3d+2a+3d+2a+3d+2a+3d=2s That is (2a+3d.n, or a+a+3 d.n, or, since a+3 d=z) a+z.n n a+z.n=2s, whence s=(- =) a+z. (THEOR. 5.) From this equation are derived a= 28 a+z 2 s n 2 (THEOR. 8.) Also by equating the values of z in 28 and n= S theorems 1 and 7, (viz, a+n—1.d= a,) we obtain a= n n 16. In like manner, by equating the values of a in theorems 2 and 6, (viz. z—n—1.d=- z,) we derive z=- + .d equating the values of n in theorems 4 and 8, we have +1= whence za-d2+2 ds-d (THEOR. 17.) a= 17. Hence any three of the five quantities a, z, d, n, s, being given, the other two may be found: also if the first term ɑ=o, any theorem containing it may be expressed in a simpler manner. 18. The following is a synopsis of the whole doctrine of arithmetical progression, wherein all the theorems above derived are brought into one view. EXAMPLES.-1. In an arithmetical progression, the first term is 3, the number of terms 50, and the common difference 2: what is the last term, and the sum of the series ? sum. Here a=3, n=50, d=2. Whence, theor. 1. z=3+50—1x2=101=the last term. 2. Given the first term 3, the last term 101, and the number of terms 50; to find the common difference and the sum of the series? 3. The first term is 3, the common difference 2, and the last term 101; required the number of terms, and the sum ? 4. The first term is 3, the number of terms 50, and the sum of the series 2600, to find the last term, and difference? 5. Given the first term 5, the last term 41, and the sum of the series 299, to find the number of terms, and the common difference? Ans. by theor. 8. n=13, and by theor. 19. d=3. Ans. by 6. Given the first term 4, the common difference 7, and the sum 355, to find the last term, and number of terms? theor. 17. z=67, and by theor. 12. n=10. 7. The last term is 67, the difference 7, and the number of terms 10, being given, to find the first term and sum? Ans. by theor. 2. a=4, and by theor. 15. s=355. 8. Let the common difference 3, the number of terms 13, and the sum 299 be given, to find the first and last terms? Ans. by theor. 9. a=5, and by theor. 13. z=41. 9. Let the last term 67, the number of terms 10, and the sum 355, be given, to find the first term and difference? Ans. by theor. 6. a=4, and by theor. 14. d=7. 10. If the last term be 9, the difference 1, and the sum 44, required the first term, and number of terms? Ans. by theor. 18. a=2, and by theor. 16. n=8. 11. The first term 0, the last term 15, and the number of terms 6, being given, to determine the difference and sum? Ans. by theor. 23. d=3, and by theor. 24. s=45. 12. Bought 100 rabbits, and gave for the first 6d. and for the last 34d. what did they cost? Ans. 81. 6s. 8d. 13. A labourer earned 3d. the first day, 8d. the second, 13d. the third, and so on, till on the last day he earned 4s. 10d. how long did the work? Ans. 12 days. 14. There are 8 equidifferent numbers, the least is 4, and the greatest 32; what are the numbers? Ans. 4, 8, 12, 16, 20, 24, 28, and 32. 15. A man paid 1000l. at 12 equidifferent payments, the first was 101.-what was the second, and the last? Ans. the second 231. 68. 8d. the last 1661. 13s. 4d. 16. A trader cleared 50l. the first year, and for 20 years he cleared regularly every year 51. more than he did the preceding; what did he gain in the last year, and what was the sum of his gains? 17. The sum of a series, consisting of 100 terms, and beginning with a cipher, is 120; required the common difference, and last term? 19. PROBLEMS EXERCISING ARITHMETICAL PROGRESSION. 1. To find three numbers in Arithmetical Progression, the common difference of which is 6, and product 35 ? Let the three numbers be x- ·6, x, and x+6 respectively. Then by the problem, (x-6.x.x+6=) x3-36x=35, or x3-36x -35=0; this equation divided by x+1, gives (x2—x—35=0, or) |