Question

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

• A. $7$ only
• B. $14$ only
• C. $7$ or $14$
• D. None of these

Answer 1

The correct option is C $7$ or $14$

Here, coefficient of $T_5$ will be ${}^n{C}_4,$ that of $T_6$ will be ${}^n{C}_5$ and that of $T_7$ will be ${}^n{C}_6.$
Since, it is given they are in A.P.
$\therefore 2^n{C}_5={}^n{C}_4{+}^n{C}_6$
Hence,
$\Rightarrow 2\left[\frac{n!}{(n-5)!5!}\right]=[\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}]$
$\Rightarrow 2\left[\frac{1}{(n-5)5}\right]=[\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}]$
$\Rightarrow 12n-48=30+n^2-9n+20\Rightarrow n^2-21n+98=0$
$\Rightarrow n=7$ or $n=14$