Question

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

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

Answer 1

Answer: (C)

7 or 14

For the expansion of $(1+x{)}^n$,
$T_5=T_{4+1}{=}^n{C}_4{x}^4$
$T_6=T_{5+1}{=}^n{C}_5{x}^5$
$T_7=T_{6+1}{=}^n{C}_6{x}^6$
So, the coefficients of $5^{th},6^{th}$ and $7^{th}$ terms are ${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$ respectively.
Since, ${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$ are in AP
${\therefore }^n{C}_4{+}^n{C}_6=2{.}^n{C}_5$
$\frac{n!}{4!(n-4)!}+\frac{n!}{6!(n-6)!}=\frac{2n!}{(n-5)!5!}$
$\Rightarrow 30+(n-5)(n-4)=2\times 6(n-4)$
$\Rightarrow n^2-21n+98=0$
$\Rightarrow (n-7)(n-14)=0$
$\Rightarrow n=7,14$