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Given an arithmetic progression -8, -3, 2,.... Determine three consecutive terms in this progression which sum up to 111.​

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Answer:

  • 9th, 10th and 11th terms; 32, 37 and 42

=============================

Given AP:

  • -8, - 3, 2, ...

We observe that:

  • The first term is a = - 8,
  • The common difference is d = -3 - (-8) = 2 - (-3) = 5.

Let the three terms be aₙ, aₙ₊₁, aₙ₊₂.

Use nth term formula:

  • aₙ = a + (n - 1)d = - 8 + (n - 1)*5 = - 8 + 5n - 5 = 5n - 13

The following two terms are:

  • aₙ₊₁ = 5n - 13 + 5 = 5n - 8
  • aₙ₊₂ = 5n - 8 + 5 = 5n - 3

Their sum is 111, set an equation and solve for n:

  • 5n - 13 + 5n - 8 + 5n - 3 = 111
  • 15n - 24 = 111
  • 15n = 135
  • n = 135/15
  • n = 9

The three terms are 9th, 10th and 11th terms and they are:

  • 5*9 - 13 = 45 - 13 = 32
  • 32 + 5 = 37
  • 37 + 5 = 42
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