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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Sequences and Series

Let A_1, A_2, A_3, , A_ 39 be 39 arithmetic means between the...

Let A1,A2,A3,…,A39A_1, A_2, A_3, \ldots, A_{39} be 3939 arithmetic means between the numbers 5959 and 159159. Then the mean of A25,A28,A31A_{25}, A_{28}, A_{31} and A36A_{36} is equal to :

Options

  1. A129129
  2. B136136
  3. C131.50131.50
  4. D134134

Correct answer

D. 134134

Step-by-step solution

The sequence 59,A1,A2,…,A39,15959, A_1, A_2, \ldots, A_{39}, 159 forms an arithmetic progression with 4141 terms.

The common difference dd is given by d=159−5939+1=10040=2.5d = \dfrac{159 - 59}{39 + 1} = \dfrac{100}{40} = 2.5

The kk-th arithmetic mean is Ak=59+k⋅dA_k = 59 + k \cdot d

The required mean is A25+A28+A31+A364\dfrac{A_{25} + A_{28} + A_{31} + A_{36}}{4}

A25+A28+A31+A364=(59+25d)+(59+28d)+(59+31d)+(59+36d)4\dfrac{A_{25} + A_{28} + A_{31} + A_{36}}{4} = \dfrac{(59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d)}{4}

=4×59+120d4=59+30d= \dfrac{4 \times 59 + 120d}{4} = 59 + 30d

Substituting d=2.5d = 2.5:

=59+30(2.5)=59+75=134= 59 + 30(2.5) = 59 + 75 = 134

Answer: 134134
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