Find the sum of the first nine terms of the arithmetic progression, if the difference between the seventh

Find the sum of the first nine terms of the arithmetic progression, if the difference between the seventh and third terms is 8, the product of the second and seventh terms is 75, and it is known that all terms of the progression are positive.

1. The ratios of the terms of a given arithmetic progression A (n) are known;

A7 – A3 = 8;

(A1 + 6 * D) – (A1 + 2 * D) =

4 * D = 8;

2. The denominator of the progression:

D = 8/4 = 2;

A2 * A7 = (A1 + D) * (A1 + 6 * D) =

(A1 + 2) * (A1 + 12) = A1² + 14 * A1 + 24 = 75;

A1² + 14 * A1 – 51 = 0;

A11.2 = -7 + – sqrt ((- 7) ² + 51) = -7 + – 10;

Since all members of the progression are positive;

3. The first member of the progression:

A1 = -7 + 10 = 3;

4. The required amount: S9 = (2 * A1 + D * (9-1)) * 9/2 =

(2 * 3 + 2 * 8) * 9/2 = 99.

Answer: the sum of nine members of the progression is 99.