The product of the forth and tenth terms of an arithmetic progression is 189. Dividing the eighth term by the third term gives 2 with a remainder of 3. Find the first term and the common difference of the progression.

The first term is 3, and the common difference is 2.

Step-by-step explanation:

The nth term of an arithmetic sequence is:

a = a₀ + d (n - 1)

We are given two equations:

a₄ a₁₀ = 189

a₈ = 2 a₃ + 3

Where:

a₃ = a₀ + d (3 - 1) = a₀ + 2d

a₄ = a₀ + d (4 - 1) = a₀ + 3d

a₈ = a₀ + d (8 - 1) = a₀ + 7d

a₁₀ = a₀ + d (10 - 1) = a₀ + 9d

Substituting:

(a₀ + 3d) (a₀ + 9d) = 189

a₀ + 7d = 2 (a₀ + 2d) + 3

Two equations, two variables. First, simplify:

a₀² + 12 a₀ d + 27d² = 189

0 = a₀ - 3d + 3

Solve for a₀ in the second equation and substitute into the first:

a₀ = 3d - 3

(3d - 3) ² + 12 (3d - 3) d + 27d² = 189

9d² - 18d + 9 + 36d² - 36d + 27d² = 189

72d² - 54d - 180 = 0

4d² - 3d - 10 = 0

4d² - 3d - 10 = 0

(d - 2) (4d + 5) = 0

d = - 1.25 or 2

a₀ = 3d - 3

a₀ = - 6.75 or 3

If a₀ = - 6.75 and d = - 1.25:

a₄ a₁₀ = - 10.5 * - 18 = 189

a₈ / a₃ = - 15.5 / - 9.25 = 1 R 6.25

Extraneous solution

If a₀ = 3 and d = 2:

a₄ a₁₀ = 9 * 21 = 189

a₈ / a₃ = 17 / 7 = 2 R 3

The first term is 3, and the common difference is 2.