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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Mathematics question on Sequences and Series

If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n are in AP and a1+a4+a7+…+a16=114a_1 + a_4 + a_7 + \ldots + a_{16} = 114, then a1+a6+a11+a16a_1 + a_6 + a_{11} + a_{16} is equal to
  1. A.
    64
  2. B.
    76
  3. C.
    98
  4. D.
    38
Show answer and solution

Correct option: B

Solution

The series a₁ + a₄ + a₇ + ... + a₁₆ has terms with indices 1,4,7,10,13,16 — that's 6 terms, each differing by 3d.

Using the AP sum property, the sum of 6 terms equally spaced equals 6 × (average of first and last) = 6 × (a₁ + a₁₆)/2 = 3(a₁ + a₁₆) = 114, so a₁ + a₁₆ = 38.

Now a₁ + a₆ + a₁₁ + a₁₆: these 4 terms are equally spaced (gap of 5d), and their sum = 4 × (average of first and last) = 4 × (a₁ + a₁₆)/2 = 2(a₁ + a₁₆) = 2 × 38 = 76.

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