The formula for the sum of the first n terms is (n/2)(2a + (n-1)*d)

where a = the first term
d = difference

So we first have to figure out a. (10/2)[2*a + (9*4)] = 80; a = -10

we use the same formula to figure out the sum of the first 40 terms

s = (40/2)[2*(-10) + (39*4)] = 2720

I dont really get the formula, however i have an alternative suggestion... you can think of it this way.

x
+ (x+4)
+ (x+4+4)
+ (x+4+4+4)
... and so on
= 80

So if you put it more simply this becomes: 10*x +(9*5)*4 = 80

(if 9*5 is not obvious to you try to think about how many fours there will be in all ten lines... you will end up at 45)


==> x = -10


So for the first 40 lines that means:

40*x + (39*20)*4 = 2,720

(Again 39*20 is the number of all the fours that there will be in the 40 lines. I thought about it as an average. There are 39 lines in which fours make an appereance. On average (since its always one more four than in the line before) there are 20 fours. Since average * number of terms = total you end up at 39*20.


Hope that was clear.