Mathematics · Binomial Theorem

JEE Main 2025 — 24 January, Morning Shift — Question 16

For some $n \neq 10,$ let the coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the binomial expansion of $(1+x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:

Option A:
70
Option B:
35
Correct
Option C:
20
Option D:
10

Answer: B

Step-by-step solution

(1+x)^{n+4}

{}^{n+4}C_4, {}^{n+4}C_5, {}^{n+4}C_6 \rightarrow A.P.

\Rightarrow 2 \times {}^{n+4}C_5 = {}^{n+4}C_4 + {}^{n+4}C_6

\Rightarrow 4 \times {}^{n+4}C_5 = ({}^{n+4}C_4 + {}^{n+4}C_5) + ({}^{n+4}C_5 + {}^{n+4}C_6)

\Rightarrow 4 \times {}^{n+4}C_5 = {}^{n+5}C_5 + {}^{n+5}C_6

\Rightarrow 4 \times \frac{(n+4)!}{5!(n-1)!} = \frac{(n+6)!}{6!n!}

\Rightarrow 4 = \frac{(n+6)(n+5)}{6n}

\Rightarrow n^2 + 11n + 30 = 24n

\Rightarrow n^2 - 13n + 30 = 0

\Rightarrow n = 3, 10 (\text{rejected})

\therefore n \neq 10

\therefore \text{ Largest\; binomial\; coefficient\; in\; expansion\; of\; } (1+x)^7

(\therefore n+4 = 7)

\text{is\; coeff.\; of\; middle\; term }

\Rightarrow {}^{7}C_4 = {}^{7}C_3 = 35

\text{ Ans\; Option \; (2)}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients