Question

The sixth term in the expansion of ${[\sqrt{\left\{2^{\log (10-3^x)}\right\}}+\sqrt[5]{5\left\{2^{(x-2)\log 3}\right\}}]}^m$is equal to $21$. If it is known that the binomial coefficient of the $2^{\text{nd}},3^{\text{rd}}$ and $4^{\text{th}}$ terms in the expansion represents respectively the first, third and fifth terms of an A.P. (the symbol log stands for logarithm to the base $10$) then the value of $m$ is

• A. $2$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is B $7$

The coefficient of the $2^{\text{nd}},3^{\text{rd}}$ and $4^{\text{th}}$ terms in the expansion are ${}^m{C}_1,{}^m{C}_2$ and ${}^m{C}_3,$ are $1^{st},3^{rd}$ and $5^{th}$ term of an A.P., so
$2{}^m{C}_2={}^m{C}_1+{}^m{C}_3$
$\Rightarrow \frac{2m(m-1)}{2!}=m+\frac{m(m-1)(m-2)}{3!}$
$\Rightarrow m(m^2-9m+14)=0$
$\Rightarrow m(m-2)(m-7)=0$
$\Rightarrow m=7$
($\because \text{for}m=0\text{or}2,$ ${}^m{C}_3$ will not exists)