- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
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 :
A
$14$
B
$21$
C
$28$
D
$7$
(JEE MAIN-2024)
Solution
$ \text { Coeff. of } x^4={ }^n C_4 $
$ \text { Coeff. of } x^5={ }^n C_5 $
$ \text { Coeff. of } x^6={ }^n C_6 $
$ { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P $
$ 2 . C_5={ }^n C_4+{ }^n 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}=\frac{n-r+1}{r}\right\} $
$ 2=\frac{5}{n-4}+\frac{n-5}{6} $
$ 12(n-4)=30+n^2-9 n+20 $
$ n^2-21 n+98=0 $
$ (n-14)(n-7)=0 $
$ n_{\max }=14 \quad n_{\min }=7$
Standard 11
Mathematics
