An arithmetic progression has 12 terms. The sum of the last six terms is three times the sum of the first 5 terms. Find the ratio of the sixth term to the fourth term.

Answer 1

S_n = sum of first n terms = (n/2)(a₁ + a_n)

So, we have S₁₂ - S₆ = 3S₅

Since S₆ = S₅ + a₆, we get S₁₂ = 4S₅ + a₆

So, (12/2)[a₁ + a₁₂] = (20/2)[a₁ + a₅] + a₁ + 5d

6(a₁ + a₁ + 11d) = 10(a₁ + a₁ + 4d) + a₁ + 5d

12a₁ + 66d = 21a₁ + 45d

So, 9a₁ = 21d. Therefore, d = (3/7)a₁

Ratio of the 6th term to the 4th term = (a₁ + 5d) / (a₁ + 3d) = (22/7)a₁ / [(16/7)a₁] = 22/16 = 11/8