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

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Dis the solution here