The sixth term of the arithmetic progression is 11 and the twelfth is -19. Find the sum

The sixth term of the arithmetic progression is 11 and the twelfth is -19. Find the sum of the first twelve terms progressively.

Let’s use the formula for the nth term of the arithmetic progression, which looks like this: an = a1 + d · (n – 1). Let’s substitute in it everything that we know first for the sixth term, and then for the twelfth term of the progression:
11 = a1 + (6 – 1) d = a1 + 5d
– 19 = a1 + (12 – 10) d = a1 + 11d
Now, subtract the upper expression from the lower expression to get rid of one of the variables:
it turns out: – 19 – 11 = 11d – 5d;
– 30 = 6d;
d = – 5.
Now we will find the first term of the arithmetic progression, substituting the found in any of the original formulas: 11 = a + 5 * (-5);
11 = a1 – 25;
a1 = 36.
The sum of the arithmetic progression is found by the formula: Sn = (a1 + an) * n / 2
S12 = (36 – 19) * 12/2 = 17 * 6 = 102.