Find the number of members of an arithmetic progression in which the sum of all 6 terms is 112

Find the number of members of an arithmetic progression in which the sum of all 6 terms is 112, the product of the second term by the difference of the progression is 30, and the sum of the third and fifth terms is 32. Write three terms of this progression.

1. Let an be an arithmetic progression. Then the sum of all members is:

Sn = 112;
(a1 + (n – 1) d) n / 2 = 112;
(a1 + (n – 1) d) n = 224. (1)
2. Let’s compose the system:

{a2d = 30;
{a3 + a5 = 32;
{(a1 + d) d = 30;
{a1 + 2d + a1 + 4d = 32;
{(a1 + d) d = 30;
{2a1 + 6d = 32;
{(a1 + d) d = 30;
{a1 + 3d = 16;
{(16 – 3d + d) d = 30;
{a1 = 16 – 3d;
{(16 – 2d) d = 30;
{a1 = 16 – 3d;
{16d – 2d ^ 2 = 30;
{a1 = 16 – 3d;
{8d – d ^ 2 = 15;
{a1 = 16 – 3d;
{d ^ 2 – 8d + 15 = 0;
{a1 = 16 – 3d;
D / 4 = 4 ^ 2 – 15 = 1;
d = 4 ± 1;
a) d = 4 – 1 = 3;

a1 = 16 – 3d = 16 – 3 * 3 = 7;

(7 + (n – 1) 3) n = 224;
(7 + 3n – 3) n = 224;
(3n + 4) n = 224;
3n ^ 2 + 4n – 224 = 0;
D1 / 4 = 2 ^ 2 + 3 * 224 = 4 + 672 = 676 = 26 ^ 2;
n = (-2 ± 26) / 3;
n1 = -28/3, does not fit;
n2 = 24/3 = 8.
b) d = 4 + 1 = 5;

a1 = 16 – 3d = 16 – 3 * 5 = 1;

(1 + (n – 1) 5) n = 224;
(5n – 4) n = 224;
5n ^ 2 – 4n – 224 = 0;
D2 / 4 = 2 ^ 2 + 5 * 224 = 4 + 1120 = 1124;
√D2 / 4 – not an integer, does not fit.
3. The only solution:

d = 3;
a1 = 7;
n = 8;
The first three members of the progression:

a1 = 7;
a2 = 7 + 3 = 10;
a3 = 10 + 3 = 13.
Answer:

1) n = 8;
2) 7, 10, 13.