If the coefficients of x 4 , x 5 and x 6 in the expansion of (1+x) n are in the arithmetic progression, then the maximum value of n is :

Mathematics · Binomial Theorem

JEE Main 2024 — 4 April, Shift 2 — Question 13

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of $(1+x)^{\mathrm{n}}$ are in the arithmetic progression, then the maximum value of n is :

Option A:
14
Correct
Option B:
21
Option C:
28
Option D:
7

Answer: A

Step-by-step solution

Coeff. of $x^{4}={ }^{n} C_{4}$

Coeff. of $x^{5}={ }^{n} C_{5}$

Coeff. of $x^{6}={ }^{n} C_{6}$

${ }^{\mathrm{n}} \mathrm{C}_{4},{ }^{\mathrm{n}} \mathrm{C}_{5},{ }^{\mathrm{n}} \mathrm{C}_{6} \ldots . \mathrm{AP}$

${ }^{\mathrm{n}} \mathrm{C}_{5}={ }^{\mathrm{n}} \mathrm{C}_{4}+{ }^{\mathrm{n}} \mathrm{C}_{6}$

$2=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C_{6}}{{ }^{n} C_{5}} \quad\left\{\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\right\}$

$2=\frac{5}{\mathrm{n}-4}+\frac{\mathrm{n}-5}{6}$

$12(n-4)=30+n^{2}-9 n+20$

$\mathrm{n}^{2}-21 \mathrm{n}+98=0$

$(\mathrm{n}-14)(\mathrm{n}-7)=0$

$\mathrm{n}_{\max }=14 \quad \mathrm{n}_{\min }=7$

Answer key and solution verified before publishing.

Practise Binomial Theorem

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients