Find the sum of the first seven terms of an arithmetic progression, the product of the third and fifth terms

Find the sum of the first seven terms of an arithmetic progression, the product of the third and fifth terms of which is equal to the second term, and the sum of the first and eighth terms is 2.

1. For a given arithmetic progression A (n), its terms meet the following conditions:

A3 * A5 = A2;

A1 + A8 = 2;

2. Let’s calculate the first term and the difference of the progression: A1 and D;

A1 + A8 = A1 + (A1 +7 * D) = 2 * A1 + 7 * D = 2;

A1 = (2 – 7 * D) / 2;

A3 * A5 = A2;

(A1 + 2 * D) * (A1 + 4 * D) = A1 + D;

A1² + 6 * D * A1 + 8 * D² = A1 + D;

3. Substitute A1 = (2 – 7 * D) / 2:

3 * D * (2 – D) = 0;

Since D> 0,

2 – D = 0;

D = 2;

A1 = (2 – 7 * 2) / 2 = -12 / 2 = -6;

A7 = A1 + 6 * d = -6 + 12 = 6;

4. The sum of the first seven members of the progression:

S7 = (A1 + A7) * 7/2 = (-6 + 6) * 7/2 = 0.

Answer: the sum of the first seven members of the progression A (n) is equal to zero.