Question

The $6^{\text{th}}$ (from the beginning) term in the expansion of the ${[\sqrt{2^{\log (10-3^x)}}+\sqrt[5]{52^{(x-2)\log 3}}]}^m$
is equal to $21$. It is known that the binomial coefficient of the $2^{\text{nd}},3^{\text{rd}}$ and $4^{\text{th}}$ term in the expansion are in an $A.P.$, then the value of $x$ is/are

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A), (C)

The correct options are
A $0$
C $2$

The coefficients are
${}^m{C}_1,{}^m{C}_2\text{and}{}^m{C}_3$
They are $T_2,T_3,T_4$ terms of an $A.P.$
$\therefore 2\cdot {}^m{C}_2={}^m{C}_1+{}^m{C}_3\Rightarrow m(m-1)=m+\frac{m(m-1)(m-2)}{6}\Rightarrow m^2-9m+14=0[\because m\ne 0]\Rightarrow m=2,7$
There should be minimum $6$ terms in the expasion, so
$m\ne 2,m=7$
Now,
$\Rightarrow T_6={}^7{C}_5{\left[\sqrt{2^{\log (10-3^x)}}\right]}^{7-5}\cdot {\left[\sqrt[5]{52^{(x-2)\log 3}}\right]}^5\Rightarrow 21=21\cdot 2^{\log (10-3^x)}\cdot 2^{(x-2)\log 3}\Rightarrow 1=2^{\log (10-3^x)+\log 3^{x-2}}\Rightarrow 0=\log (10-3^x)+\log 3^{x-2}\Rightarrow (10-3^x)\left(3^{x-2}\right)=1\Rightarrow (10-3^x)3^x=9\Rightarrow 3^{2x}-10\cdot 3^x+9=0\Rightarrow (3^x-9)(3^x-1)=0\Rightarrow 3^x=9,1\Rightarrow x=2,0$