Question

Match the following by appropriately matching the lists based on the information given in $\text{Column I}$ and $\text{Column II}$.

$\begin{array}{cc}ColumnI& ColumnII\\ \text{a. The coefficient of two consecutive terms in the}& \text{p.}9\\ \text{expansion of}(1+x{)}^n\text{will be equal, then}n\text{can be}& \\ \text{b. If}{15}^n+{23}^n\text{is divisible by}19,\text{then}n\text{can be}& \text{q.}10\\ \text{c. If the coefficients of}{T}_r,T_{r+1},T_{r+2}\text{terms of}& \text{r.}11\\ (1+x{)}^{14}\text{are in A.P., then}r\text{is less than}& \end{array}$

• A. $a-p,r;b-p,r;c-q,r$
• B. $a-p,q;b-p,q;c-p,q,r$
• C. $a-p,q,r;b-p,q,r;c-p,q,r$
• D. $a-q,r;b-q,r;c-p,r$

Answer 1

The correct option is A $a-p,r;b-p,r;c-q,r$

$\text{a.}$
Since, the coefficients of consecutive terms is same,
$\therefore {}^n{C}_r={}^n{C}_{r+1}$ for some $r$
$\therefore \frac{n!}{(n-r)!\times r!}=\frac{n!}{(n-r-1)!\times (r+1)!}$
or, $\frac{n!}{(n-r)\times (n-r-1)!\times r!}=\frac{n!}{(n-r-1)!\times (r+1)r!}$
or, $r+1=n-r$
or, $n=2r+1$
Hence, $n$ is odd.

$\text{b.}$
$E={15}^n+{23}^n$
$\Rightarrow E=(19-4{)}^n+(19+4{)}^n$
$\Rightarrow E=2\left[{}^n{C}_0\right(19{)}^n+{}^n{C}_2(19{)}^{n-2}{4}^2+\dots +{}^n{C}_n{4}^n]$, when $n$ is even,

or $E=2\left[{}^n{C}_0\right(19{)}^n+{}^n{C}_2(19{)}^{n-2}{4}^2+\dots +{}^n{C}_{n-1}(19)\cdot 4^{n-1}]$, when $n$ is odd

Hence, $E$ is divisible by $19$ when $n$ is odd.

$\text{c.}$
$T_r={}^{14}{C}_{r-1}{x}^{r-1}$
$T_{r+1}={}^{14}{C}_r{x}^r$
$T_{r+2}={}^{14}{C}_{r+1}{x}^{r+1}$
By given condition,
$2\times {}^{14}{C}_r={}^{14}{C}_{r-1}+{}^{14}{C}_{r+1}$
$\Rightarrow 2=\frac{{}^{14}{C}_{r-1}}{{}^{14}{C}_r}+\frac{{}^{14}{C}_{r+1}}{{}^{14}{C}_r}$
$\Rightarrow 2=\frac{r}{14-r+1}+\frac{14-(r+1)+1}{r+1}$
$\Rightarrow 2=\frac{r}{15-r}+\frac{14-r}{r+1}$
$\Rightarrow r^2-14r+45=0$
$\Rightarrow r=5,9$