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? Here a=3, z=101, n=50. 101-3 Whence, theor. 3. d=- =2=the common difference. 50-1 50 And theor. 5. s=3+101 x=2600= the sum. 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. 6. Given the first term 4, the common difference 7, and the sum 355, to find the last term, and number of terms? Ans. by 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 he 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 10-what was the second, and the last? Ans. the second 231, 68. Sd, the last 1661. 13s. 4d. 16. A trader cleared 501, 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—36 x -350; this equation divided by x+1, gives (x-x-35=0, or) r2-x=35; which resolved, we have x=35.25+.5, whence x-6=35.25 +5.5, and x+6=√√/35.25 +6.5: the numbers therefore are .43717, 6.43717,and 12.43717, nearly. 2. An artist proposed to work as many days at 3 shillings per day, as he had shillings in his pocket; at the end of the time having received his hire, and spent nothing, he finds himself worth 44 shillings; what sum did he begin with? Let x=his number of shillings at first, whence also x=the number of days he worked: we have therefore here given the first term x, the common difference 3, and the number of terms x+1, in an arithmetical progression, to find the last term; now by theor. 1. (z=a+n−1.d, or) 44=x+x+1−1x3, that is, 4x= 44, whence x=11 shillings=the sum he began with. 3. To find three numbers in arithmetical progression, such, that their sum may be 12, and the sum of their squares 56? Let x=the common difference, 3 s=(12) the sum, then will s=the middle number, s-x=the less extreme, and s+x=the greater extreme, also let a=56; then by the problem, (s—x)2+ s2+s+x}2=) 3s2+2x2=a, whence 2x2-a-3so, and x= 3 s2 56-48 a 2 2 =2; therefore s=4, s-x=2, and s+x=6, that is, 2, 4, and 6, are the numbers required. 4. To find four numbers in arithmetical progression, whereof the product of the extremes is 52, and that of the means 70? Let x=the less extreme, y=the common difference; then will x, x+y, x+2y, and x+3y, represent the progression. Let a= 52, b=70, then by the problem (x.x+3y=) x2+3xy=a, and (x+y.x+2y=) x2 + 3 xy +2 y2=b; from the latter equation subtract the former, and 2y=b-a, whence y=√ stitute this value for y in the first equation, and it becomes x2+9x b- -a =a; completing the square, &c. we obtain x=√a+ wherefore 4, 7, 10, and 13, are the numbers required. =3; sub 81 9 :4: 5. The sum of six numbers in arithmetical progression is 48, and if the common difference d be multiplied into the less extreme, the product equals the number of terms; required the terms of the progression ? |