Find the sum of the first 12 terms of the arithmetic progression if a1 = -3, a3 * a7 = 24

Let’s write down the relations between the first, third and seventh terms of the arithmetic progression and compose the equation.

a3 = a1 + 2d;
a7 = a1 + 6d;

(a1 + 2d) (a1 + 6d) = 24;
(- 3 + 2d) (- 3 + 6d) = 24;

9 – 6d – 18d + 12d2 = 24;

12d2 – 24d – 15 = 0.

4d2 – 8d – 5 = 0.

D = 8 * 8 + 4 * 4 * 5 = 64 + 80 = 144 = 122.

d1 = (8 + 12) / 4 * 2 = 10/4 = 2.5.
d2 = (8 – 12) / 4 * 2 = – 2.

Let’s find the sum of the first 12 members of the sequence using the formula for both possible values of d.

S12 = (2a1 + (12 – 1) d) 12/2 = (2 * (- 3) + 11 * 2.5) * 6 = (27.5 – 6) * 6 = 21.5 * 6 = 129 …

S12 = (2a1 + (12 – 1) d) 12/2 = (2 * (- 3) – 22) * 6 = – 28 * 6 = 48.

Answer: possible values of the sum S12 = 129 and S12 = 48 for two variants of an increasing and decreasing progression.