"In mathematics, an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13... is an arithmetic progression with common difference 2. If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence is given by: an = a1 + (n - 1) * d " Source and further information: http://en.wikipedia.org/wiki/Arithmetic_progression Here we have two arithmetic progressions with: an = a1 + (n - 1) * da bn = b1 + (n - 1) * db "The sum of the components of an arithmetic progression is called an arithmetic series. Sn = n*[2*a1 + (n -1)*d] " (Same source) The sums of the first n terms for the "a" progression is: San = n*[2*a1 + (n -1)*da] The sums of the first n terms for the "b" progression is: Sbn = n*[2*b1 + (n -1)*db] The sums of the first n terms are in the ratio: San : Sbn = [2*a1 + (n -1)*da] : [2*b1 + (n -1)*db] = [da*n + (2*a1 - da)] : [db*n + (2*b1 - db)] From this and the question, we get for instance: da = 7 2*a1 - da = 1 a1 = 8/2 = 4 db = 4 2*b1 - db = 27 b1 = 31/2 = 15.5 Further solutions would be possible where all quantities would be multiplied by the same factor. The ratio of the nth term of the arithmetic progressions is: [a1 + (n - 1) * da] : [b1 + (n - 1) * db] = [4 + (n - 1) * 7] : [15.5 + (n - 1) * 4] = (7*n - 3):(4*n + 11.5) (this applies to all solutions, because the factors disappear by the division) For n = 11 we get: (7*11 - 3):(4*11 + 11.5) = (77 - 3):(44 + 11.5) = 74/55.5 = (37*2)/(37*3/2) = 4/3 = 1.333

The sum of a arithmetic sequence: a, a+d, a+2d, a+3d, ..., a+(n-1)d is na + n(n-1)d/2 (use the rule "average of first and last multiplied by number of entries") If we were to divide these, one factor of n would disappear and wed' be left with a + (n-1)d/2 or (d/2) n + (a-(d/2)) But this is just the same form as the ratios in the question. (something times n + something) So for the first progression make d=14 and a=8 And for the second progression make d=8 and a=31 The 11th term of each progression is a+10d So the terms are: 148 for the first progression. and 111 for the second progression. The ratio is 148:111 Check: sequence 1: 8, 22, 36, 50, 64, 78, 92, 106, 120, 134, 148 sums: 8, 30, 66, 116, 180, 258, 350, 456, 576, 710, 858 sequence 2: 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111 sums: 31, 70, 117, 172, 235, 306, 385, 472, 567, 670, 781 ratios of sums: 8:31, 15:35, 22:39, 29:43, 36:47, 43:51, 50:55, 57:59, 64:63, 71:67, 78:71