Question

Arrange the values of $n$ in ascending order
A : If the term independent of $x$ in the expansion of ${\left(\sqrt{x}-\frac{n}{x^2}\right)}^{10}$ is $405$
B : If the fourth term in the expansion of ${\left(\frac{1}{n}+n^{{\log }_{n}10}\right)}^5$ is $1000$, ( $n<10$)
C : In the binomial expansion of $(1+x{)}^n$ the coefficients of $5^{th},6^{th}$ and $7^{th}$ terms are in A.P.

• A. $A,B,C$
• B. $B,A,C$
• C. $A,C,B$
• D. $C,A,B$

Answer 1

Answer: (B)

$B,A,C$

A: Given: $(\sqrt{x}-n{x}^{-2}{)}^{10}$

$T_{r+1}=(-1{)}^r{}^{10}{C}_r{x}^{(5-5r/2)}{n}^r$ ...(1)

Term independent of $x$ implies
$5-\frac{5r}{2}=0$

Thus $r=2$

Substituting in equation (1), we get
$T_3={}^{10}{C}_2{n}^2=405$
$\Rightarrow 45n^2=405$
$\Rightarrow n^2=9$
$\Rightarrow n=3$ ($n$ is $\ne -3$ as $n$ is a positive integer)

B: Given: $(n^{-1}+n^{lo{g}_{n}10}{)}^5$
$=(n^{-1}+10{)}^5$
$T_{r+1}={}^5{C}_r{n}^{5-r}{10}^r$ ...(2)
It is given that the third term is $1000$
Substituting in equation (2), we get
$T_4={}^5{C}_2{n}^{-2}{10}^3=1000$
$\Rightarrow 10\left(n^{-2}1000\right)=1000$
$\Rightarrow n^{-2}=10$
$\Rightarrow n=\frac{1}{\sqrt{10}}$

C: For the terms to be in arithmetic progression
$(n-2r{)}^2=n+2$ ...(3)
Let us consider ${}^n{C}_{r-1},{}^n{C}_r,{}^n{C}_{r+1}$ be in A.P.
Hence, $T_r,T_{r+1},T_{r+2}$ are in A.P.
Here it is given that, the $5^{th},6^{th}$ and $7^{th}$ terms are in A.P.
Hence, $r=5$
Substituting in equation (3), we get
$(n-10{)}^2=n+2$
$\Rightarrow n^2-21n+98=0$
$\Rightarrow (n-7)(n-14)=0$
Hence, $n=7,14$
Hence, the values of $n$ in ascending order is B,A,C.

Option B.