The difference between the seventh and second terms of the arithmetic progression is 15, and the product

The difference between the seventh and second terms of the arithmetic progression is 15, and the product of the second and seventh terms is 250. How many members of this progression must be taken to get a total of 87?

Let’s write the equations:
a7 – a2 = (a1 + 6d) – (a1 + d) = 5 d = 15;

d = 3.

a7 a2 = (a1 + 6d) (a1 + d) = (a1 + 18) (a1 + 3) = 250;

a12 + 21 a1 – 196 = 0;

a1 = (- 21 ± √ (441 + 4 * 196)) / 2 = (- 21 ± 35) / 2.

a) a1 = 7.

Sn = n (a1 + an) / 2 = n (a1 + a1 + 3 (n – 1)) / 2 = 87;

n (3 n + 11) = 174;

3 n2 + 11 n – 174 = 0;

n = (-11 ± √ (121 + 12 * 174)) / 6 = (-11 ± 47) / 6

n = 6.

The second value n = – 8 is not taken into account.

b) a1 = -28.

Sn = n (a1 + an) / 2 = n (a1 + a1 + 3 (n – 1)) / 2 = 87;

n (- 56 + 3 (n – 1)) / 2 = 87;

3 n2 – 59 n – 174 = 0;

The discriminant of this equation is 5569 is not the square of an integer, and therefore n is not an integer.

In this way:

a1 = 7, d = 3; n = 6.