a quick way to solve this, is like this:

Lets say you have series of consequtive numbers:

A1 A2 A3 A4 ..... An

just to make things simple, lets say n is even (although it doesnt really matter)

its obvious that A1 + An = A2 + An-1 = A3 + An-2 and so on ... this is the definition of arithmetic progression

so you have n/2 couples with the same sum which is A1+An, the sum of the series will be n/2(An+A1) - Disclaimer: just to be accurate, im referring to n as the number of elements in the series, A1 doesnt necessarily has to be 1, and in this context 'n' is not necessarily the 'n' in An - i can clarify it if its not clear
i assume you'd understand what i mean ...

now for our private case:

our series is :

2 4 6 ..... n

so we actually have n/4 couples with a sum n+2 ! (n/4 because we first divide by 2 by creating a couple, but we have n/2 elements in the first place)

so: n/4 (n+2) = 79*80 => n = 158

This explenation was long, but if you remember it, you can solve these kind of questions in a flink of an eye

so we actually have n/4 couples with a sum n+2 !

I did not get why sum is n+2!. : I'm convinced with n/4 couples.

By the way cool approach. thanks.

Hey got it 'n' is the number of terms, then n/2 couples and the sum will be n/2 (A1+ An) which in this case n/4 * (n+2). Great.

thanks a lot. THat was real quickie.