To sum up the terms of this arithmetic sequence:
a + (a+d) + (a+2d) + (a+3d) + ...
use this formula:
Let's see why the formula works, because we get to use an interesting "trick" which is worth knowing.
First, we will call the whole sum "S":
Next, rewrite S in reverse order:
Now add those two, term by term:
S | = | a | + | (a+d) | + | ... | + | (a + (n-2)d) | + | (a + (n-1)d) |
S | = | (a + (n-1)d) | + | (a + (n-2)d) | + | ... | + | (a + d) | + | a |
2S | = | (2a + (n-1)d) | + | (2a + (n-1)d) | + | ... | + | (2a + (n-1)d) | + | (2a + (n-1)d) |
Each term is the same! And there are "n" of them so ...
Now, just divide by 2 and we get:
S = (n/2) × (2a + (n−1)d)
Which is our formula: