Multiple choice general knowledge math & puzzles

In an arithmetic series, the 3rd term is equal to 114 and the last term is equal to -27. The sum of all terms in the series is equal to 2325. What are the last three terms in the series?

  1. -7 , -17 , -27

  2. -25 , -26 , -27

  3. 27 , 0 , -27

  4. -21 , -24 , -27
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the arithmetic series formulas: nth term aₙ = a₁ + (n-1)d and sum Sₙ = n(a₁ + aₙ)/2. Given a₃ = 114, aₙ = -27, and S = 2325. From a₃ = a₁ + 2d = 114, we get a₁ = 114 - 2d. From S = n(a₁ + aₙ)/2 = n((114-2d) + (-27))/2 = n(87 - 2d)/2 = 2325. From aₙ = a₁ + (n-1)d: -27 = (114-2d) + (n-1)d = 114 - 2d + nd - d = 114 + nd - 3d. Solving gives d = -3, n = 60, a₁ = 120. The last three terms are: 58th = a₁ + 57d = 120 + 57(-3) = -51, 59th = 120 + 58(-3) = -54, 60th = 120 + 59(-3) = -57. Wait - this doesn't match option D. Let me recalculate: -27 = a₁ + (n-1)(-3), so a₁ = -27 + 3(n-1). Also a₁ = 114 - 2(-3) = 120. So 120 = -27 + 3n - 3, so 3n = 150, n = 50. Last three: 48th = 120 + 47(-3) = -21, 49th = 120 + 48(-3) = -24, 50th = 120 + 49(-3) = -27. This matches option D.

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