AP Calc BC Sequences question...?

Question

If a1+a2+a3+...+a80=2012 and the sequence has a common difference of 5, what is the sum of a2+a4+a6+...+a80?

I cannot figure out how to do this! Any help is appreciated (and no, I'm not allowed to just straight up add all the numbers :P).

Answer 1

Using S_n = (n/2)[2a + (n-1)d] with sum = 2012, n = 80, d = 5, you can solve for the value of a (first term).

 

I got a = -172.35   CHECK MY WORK before proceeding.

 

Then the sum of forty terms with even subscripts is a new arithmetic series, n = 40, a = -172.35+5, d = 10.

 

You can take it from here!