Mathematics · Sequence and Series

JEE Main 2026 — 5 April, Evening Shift — Question 30

Let $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots \ldots . ., \mathrm{A}_{39}$ be $39$ arithmetic means between the numbers $59$ and $159.$ Then the mean of $\mathrm{A}_{25}, \mathrm{~A}_{28}, \mathrm{~A}_{31}$ and $\mathrm{A}_{36}$ is equal to :

Option A:
$129$
Option B:
$136$
Option C:
$131.5$
Option D:
$134$
Correct

Answer: D

Step-by-step solution

$59 \mathrm{~A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3} \ldots \mathrm{~A}_{39} 159$ $\mathrm{d}=\frac{159-59}{39+1}=\frac{5}{2}$ Now, $\mathrm{A}_{25}=59+25 \mathrm{~d}=121.5$

\begin{aligned} & A_{28}=59+28 d=129 \\& A_{31}=59+31 d=136.5 \\& A_{36}=59+36 d=149 \end{aligned}

Mean of

\begin{aligned} & A_{25}, A_{28}, A_{31}, A_{36}=\frac{121.5+129+136.5+149}{4} \\& =\frac{536}{4}=134 \end{aligned}

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

Answer: D

Step-by-step solution

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