In any artihmetic sequence, consecutive terms differ by a fixed constant <em>c</em>. So given the first term <em>a</em>, the second term is <em>a</em> + <em>c</em>, the third terms is <em>a</em> + 2<em>c</em>, and so on, up to the <em>n</em>-th term <em>a</em> + (<em>n</em> - 1)<em>c</em>.
If the 15th term is 40, then
40 = -12 + (15 - 1) <em>c</em> ==> <em>c</em> = 52/14 = 26/7
We can then write the <em>n</em>-th term as
-12 + (<em>n</em> - 1) 26/7 = (26<em>n</em> - 110)/7
The sum of the first 15 terms is then
Another way to compute the sum: let <em>S</em> denote the sum,
<em>S</em> = -12 - 58/7 - 32/7 + … + 228/7 + 254/7 + 40
Reverse the order of terms:
<em>S*</em> = 40 + 254/7 + 228/7 + … - 32/7 - 58/7 - 12
Notice that adding up terms in the same position gives the same result,
-12 + 40 = 28
-58/7 + 254/7 = 28
-32/7 + 228/7 = 28
so that
<em>S</em> + <em>S*</em> = 2<em>S</em> = 28 + 28 + 28 + … + 28 + 28 + 28
There are 15 terms in the sum, so
2<em>S</em> = 15×28 ==> <em>S</em> = 15×28/2 = 210