JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

Let a₁, a₂, ..., an be a given A.P. whose
common difference is an integer and
Sₙ = a₁ + a₂ + .... + aₙ. If a₁ = 1, aₙ = 300 and 15 ≤ n ≤ 50, then
the ordered pair (Sₙ₋₄, a n–4 ) is equal to:

  1. (a) (2480, 249)
  2. (b) (2480, 248)
  3. (c) (2490, 248)
  4. (d) (2490, 249)
Answer and solution

Answer (c) (2490, 248)

Solution

Maths sir · 4 steps

Integer common difference means n-1 must divide 299. Factor 299 and the range picks n for you.

  1. Use the last term. aₙ = a₁ + (n-1)d gives 1 + (n-1)d = 300, so (n-1)d = 299. Since d is an integer, n-1 must be a divisor of 299.
  2. Factor 299. 299 = 13 × 23, so its divisors are 1, 13, 23 and 299. With 15 ≤ n ≤ 50 we need 14 ≤ n-1 ≤ 49, and only n-1 = 23 fits. So n = 24 and d = 13.
  3. Find the 20th term. n - 4 = 20, so a₂₀ = 1 + 19 × 13 = 1 + 247 = 248.

    S₂₀ = 20/2(a₁ + a₂₀) = 10 × (1 + 248) = 2490

  4. So. (Sₙ₋₄, aₙ₋₄) = (2490, 248), which is option (c).

The idea to keep. When the common difference must be an integer, (n-1)d equals a fixed number. Factor that number and let the given range of n choose the right divisor.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 250 times in JEE Main since 2020. This question is from 2020.