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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Sequences and Series

Let the sum of the first nn terms of a non-constant AP a1,a2,a3,…a_1, a_2, a_3, \ldots be 50n+n(n−7)2A50n + \dfrac{n(n-7)}{2}A, where AA is a constant. If dd is the common difference of this AP, then the ordered pair (d,a50)(d, a_{50}) is equal to
  1. A.
    (A,50+46A)(A, 50 + 46A)
  2. B.
    (50,50+45A)(50, 50 + 45A)
  3. C.
    (50,50+46A)(50, 50 + 46A)
  4. D.
    (A,50+45A)(A, 50 + 45A)
Show answer and solution

Correct option: A

Solution

The nth term is aₙ = Sₙ - Sₙ₋₁.

Sₙ = 50n + n(n-7)A/2

Sₙ₋₁ = 50(n-1) + (n-1)(n-8)A/2

aₙ = 50 + A/2 · [n(n-7) - (n-1)(n-8)] = 50 + A/2 · [n²-7n - (n²-9n+8)] = 50 + A/2 · (2n-8) = 50 + A(n-4)

So aₙ = 50 + A(n-4), which is linear in n, confirming it's an AP with common difference d = A.

a₅₀ = 50 + A(50-4) = 50 + 46A.

Thus (d, a₅₀) = (A, 50 + 46A).

[JEE Main 2019 · 9 Apr · Shift I]

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