To sum up the terms of this arithmetic sequence:

a + (a+d) + (a+2d) + (a+3d) + ... 

use this formula:

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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":

S = a + (a + d) + ... + (a + (n−2)d) + (a + (n−1)d) 

Next, rewrite S in reverse order:

S = (a + (n−1)d) + (a + (n−2)d) + ... + (a + d) + a

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 ...

2S = n × (2a + (n−1)d) 

Now, just divide by 2 and we get:

S = (n/2) × (2a + (n−1)d) 

Which is our formula:

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