Question

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.

Answer 1

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

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.