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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Sequences and Series

Let a1,a2,…,a30a_1, a_2, \ldots, a_{30} be an AP, S=∑i=130aiS = \sum_{i=1}^{30} a_i and T=∑i=115a(2i−1)T = \sum_{i=1}^{15} a_{(2i-1)}. If a5=27a_5 = 27 and S−2T=75S - 2T = 75, then a10a_{10} is equal to
  1. A.
    42
  2. B.
    57
  3. C.
    52
  4. D.
    47
Show answer and solution

Correct option: C

Solution

Let the AP have first term a and common difference d.

S = (30/2)(2a + 29d) = 15(2a + 29d)

T = a₁ + a₃ + a₅ + ... + a₂₉ (odd-indexed terms)
These form an AP with first term a, common difference 2d, 15 terms.
T = (15/2)(2a + 14·2d) = 15(a + 14d)

S - 2T = 15(2a + 29d) - 2·15(a + 14d) = 15(2a + 29d - 2a - 28d) = 15d = 75
So d = 5.

a₅ = a + 4d = a + 20 = 27 → a = 7.

a₁₀ = a + 9d = 7 + 45 = 52.

[JEE Main 2019 · 9 Jan · Shift I]

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