The sum of the first n terms , Sn of a progression is given by Sn=5n-n^2 .

Question

How to find in terms of n, the sum of the first (n-1) terms , Sn-1 . Then how to find in terms of n, the nth term, Tn, of the progression and prove that the progression is an A.P.

Answer 1

S_n-1 = 5(n-1)-(n-1)^2 = 5n - 5 - n^2 + 2n - 1 = -n^2 + 7n - 6

Thus, T_n = 5n - n^2 - (-n^2 + 7n - 6) = 6 - 2n

As a check, note that this works for n = 1, 2, 3

For n=1, T_n = 6 - 2n = 6 - 2*1 = 4, and S_n = 5n - n^2 = 5*1-1^2 = 5 - 1 =4 4 = 4

For n=2, T_n = 6 - 2n = 6 - 2*2 = 2, so Sn calculated as a sum is 4 + 2 = 6 and S_n = 5n - n^2 = 5*2-2^2 = 10 - 4 = 6 6=6

For n=3, T_n = 6 - 2n = 6 - 2*3 = 0, so Sn calculated as a sum is 4 + 2 + 0 = 6 and S_n = 5n - n^2 = 5*3-3^2 = = 15 - 9 = 6 6 = 6

An A.P. is in the form T_n = cn + d for c not equal to 0. Thus, this is an A.P. (you can replace n with n-1 in the expression, but all this does is change d, which does not change the form of the expression.