The common difference of the A.P.: a 1 , a 2 , ldots, a m is 13 more than the common difference of the A.P.: b 1 , b 2 , ldots, b n . If b 31 =-277, b 43 =-385 and a 78 = 327, then a 1 is equal to

Mathematics · Sequence and Series

JEE Main 2026 — 28 January, Morning Shift — Question 16

The common difference of the A.P.: $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{m}}$ is $13$ more than the common difference of the A.P.: $\mathrm{b}_{1}, \mathrm{~b}_{2}, \ldots, \mathrm{~b}_{\mathrm{n}}$ . If $\mathrm{b}_{31}=-277, \mathrm{~b}_{43}=-385$ and $\mathrm{a}_{78}=$ 327, then $\mathrm{a}_{1}$ is equal to

Option A:
21
Option B:
24
Option C:
19
Correct
Option D:
16

Answer: C

Step-by-step solution

Let common difference of A.P.'s are $\mathrm{d}_{1} \& \mathrm{~d}_{2}$

$\therefore \mathrm{d}_{1}=13+\mathrm{d}_{2}$ $\begin{gathered} \mathrm{b}_{1}+30 \mathrm{~d}_{2}=-277 \end{gathered}$ $\begin{gathered} \mathrm{b}_{1}+42 \mathrm{~d}_{2}=-385 \end{gathered}$ By - (1) $12 \mathrm{~d}_{2}=-108$

$\mathrm{d}_{2}=-9$

$\therefore \mathrm{d}_{1}=4$

Now $\mathrm{a}_{78}=327$

$\Rightarrow \mathrm{a}_{1}+77 \mathrm{~d}_{1}=327$

$\Rightarrow \mathrm{a}_{1}+308=327$

$\mathrm{a}_{1}=19$

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

Answer: C

Step-by-step solution

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