Find the 41st term of an arithmetic sequence if the common difference is - 4 and the first term is 2.

In an arithmetic series, an = a0 + (n-1) * d that means the nth term is equal to the first term plus n-1 times the common difference. In this case we are told a3=10 and a5=16 so: a3 = a + 2*d = 10 = > a0 = 10-2d and a5 = a+4*d = 16 = > a0 = 16-4d so 10-2d = 16-4d = > 2d=6 = > d=3 now you can plug this value of d into any of those equations we got for a0 and find a0. for example a0 = 10-2 (3) = 4 so the first term is 4 the formula for the sum of the first n terms is: Sn = (n/2) * [2a0 + (n-1) * d] so S20 = (20/2) * [2*4 + (19) * (3) ] = 10 * [65] = 650