also there is no solution to this at all.
We know that the sum of an arithmetic sequence is as follows:
S_n = \frac n2 (2a_1 + (n-1)d)
and so plugging numbers in should leave you with S_n = \frac{600}{2}(2\cdot 3 + (600-1)\cdot 2) = 361200
Idk what you mean with “number of integers in intervals and sum of integers…” thing, but a solution to this question could look like what i have written above.
but this wasn’t covered in greg videos
, what was covered then? I can’t give you specific help if you don’t provide context, right?
this was series 1 video and greg never covered this formula anywhere in this video. it is really disheartening throwing in questions like this and making a topic so difficult
You can use the approaches described in “number/sum of integers in an interval”. Share your steps.
Hi Greg
I used the number of integers in the interval to which was 599-3+1 and then the sum of integers in the interval which is 599+3 and then multiplied both to get divided by 2.
But the answer was way off
How? In particular, how’s the 600th term of the sequence “599”?