For the following two (02) items: Let (6 + 10 + 14 + up to m terms ) = (1 + 3 + 5 + 7 + up to n terms ) , where m How many values of m are possible?

Step-by-step solution

The sum of the first m terms of the arithmetic progression 6, 10, 14, is given by: S_m = m 2 [2(6) + (m - 1)4] = m 2 [12 + 4m - 4] = 2m^2 + 4m The sum of the first n terms of the arithmetic progression 1, 3, 5, 7, is given by: S_n = n 2 [2(1) + (n - 1)2] = n^2 Equating the two sums, we get: 2m^2 + 4m = n^2 2(m^2 + 2m + 1) - 2 = n^2 2(m + 1)^2 - n^2 = 2 Let u = m + 1 . The equation becomes 2u^2 - n^2 = 2 . This implies that n must be an even number. Let n = 2v . Substituting this into the equation yields: 2u^2 - 4v^