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 sum of possible values of $x$ is

• A. $1$
• B. $3$
• C. $4$
• D. $2$

Answer 1

The correct option is D $2$

The coefficient of the $2{}^{nd},3^{rd}and{4}^{th}$ terms in the expansion are ${}^m{C}_1,{}^m{C}_2$ and ${}^m{C}_3,$ which are given in A.P. Hence,
$2{}^m{C}_2={}^m{C}_1+{}^m{C}_3$
$\Rightarrow \frac{2m(m-1)}{2!}=m+\frac{m(m-1)(m-2)}{3!}$
$\Rightarrow m(m^2-9m+14)=0$
$\Rightarrow m(m-2)(m-7)=0$
$\Rightarrow m=7$
($\because \text{for}m=0\text{or}2$ ${}^m{C}_3$ will not exists)
Now, $6^{th}$
term in the expansion, when $m=7$, is
$T_6{=}^7{C}_5{\left[\sqrt{2^{\log (10-3^x)}}\right]}^{7-5}\times {\left[\sqrt[5]{52^{x-2\log 3}}\right]}^5=21$
$\Rightarrow \frac{7\times 6}{2!}[2^{\log (10-3^x)}\times 2^{(x-2)\log 3}]=21$
$\Rightarrow 2^{\log (10-3^x)+(x-2)\log 3}=1=2^0$
$\Rightarrow \log (10-3^x)+(x-2)\log 3=0$
$\Rightarrow \log (10-3^x)(3{)}^{x-2}=0$
$\Rightarrow (10-3^x)\times 3^x\times 3^{-2}=1$
$\Rightarrow 10\times 3^x-(3^x{)}^2=9$
$\Rightarrow (3^x{)}^2-10\times 3^x+9=0$
$\Rightarrow (3^x-1)(3^x-9)=0$
$\Rightarrow 3^x-1=0\Rightarrow 3^x=1=3^0\Rightarrow x=0$
$\Rightarrow 3^x-9=0\Rightarrow 3^x=3^2\Rightarrow x=2$
Hence, $x=0$ or $2.$
$\therefore$ Sum of possible values of $x$ is $0+2=2$