Question

If the coefficients of $5^{th}$, $6^{th}$ and $7^{th}$ terms in the expansion of $(1+x{)}^n$ are in A.P. then $n=$

• A. $7$
• B. $14$
• C. $7or14$
• D. $8$

Answer 1

The correct option is C $7or14$

We know the binomial expansion of $(1+x{)}^n$,

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+....{+}^n{C}_{n-1}{x}^{n-1}{+}^n{C}_x{x}^n$

Given: coefficients of $5^{th},6^{th},7^{th}$ terms in the expansion are in A.P.

So, ${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$ are in A.P.

The first, second and third term of an A.P., is written as $a_2-a_1=a_3-a_2\Rightarrow 2a_2=a_3+a_1$

Substitute the values,

$2^n{C}_5{=}^n{C}_4{+}^n{C}_6$

$\Rightarrow 2=\frac{{}^n{C}_4{+}^n{C}_6}{{}^n{C}_5}$

$\Rightarrow 2=\frac{{}^n{C}_4}{{}^n{C}_5}+\frac{{}^n{C}_6}{{}^n{C}_5}$

$\Rightarrow 2=\frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}}+\frac{\frac{n!}{6!(n-6)!}}{\frac{n!}{5!(n-5)!}}$

$\Rightarrow 2=\frac{5!(n-5)!}{4!(n-4)!}+\frac{5!(n-5)!}{6!(n-6)!}$

$\Rightarrow 2=\frac{5\times 4!(n-5)!}{4!(n-4)(n-5)!}+\frac{5!(n-5)(n-6)!}{6\times 5!(n-6)!}$

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

cross multiplying we get,

$\Rightarrow 2=\frac{30+(n-4)(n-5)}{6(n-4)}$

$\Rightarrow 12n-48=30+n^2-4n-5n+20$

$\Rightarrow 12n-48=50+n^2-9n$

$\Rightarrow n^2-9n-12n+48+50=0$

$\Rightarrow n^2-21n+98=0$

$\Rightarrow n^2-14n-7n+98=0$

$\Rightarrow n(n-14)-7(n-14)=0$

$\Rightarrow (n-7)(n-14)=0$

$\Rightarrow n=7,14$