Question

The sum of all possible value(s) of $x$ for which the sixth term of ${[{\left(2^{\ln (10-3^x)}\right)}^{1/2}+{\left(2^{(x-2)\ln 3}\right)}^{1/5}]}^m$ is equal to $21$ and binomial coefficients of second, third and fourth terms are the first, third and fifth terms of an arithmetic progression, is

• A. $2$
• B. $0$
• C. $-1$
• D. $5$

Answer 1

The correct option is A $2$

Let $a=2^{\ln (10-3^x)}$ and $b=2^{(x-2)\ln 3}$
Then, $T_{r+1}={}^m{C}_r\cdot (a{)}^{(m-r)/2}\cdot (b{)}^{r/5}$
$\therefore T_6={}^m{C}_5\cdot (a{)}^{(m-5)/2}\cdot (b)=21\cdots (1)$
Also, ${}^m{C}_1,{}^m{C}_2,{}^m{C}_3$ are in A.P.
$\therefore 2\cdot {}^m{C}_2={}^m{C}_1+{}^m{C}_3$
$\Rightarrow 2\cdot \frac{m(m-1)}{2}=m+\frac{m(m-1)(m-2)}{6}$
$\Rightarrow 6(m-1)=6+(m-1)(m-2)(\because m\ne 0)$
$\Rightarrow m=7,2$(rejected)

Substituting $m=7$ in equation $\left(1\right),$ we get
${}^7{C}_5\cdot a\cdot b=21$
$\Rightarrow ab=1$
$\Rightarrow 2^{\ln (10-3^x)}\cdot 2^{(x-2)\ln 3}=1$
$\Rightarrow 2^{\ln (10-3^x)+(x-2)\ln 3}=2^0$
$\Rightarrow \ln (10-3^x)+(x-2)\ln 3=0$
$\Rightarrow \ln \left\{\right(10-3^x\left)\right(3^{(x-2)}\left)\right\}=0$
$\Rightarrow \frac{(10-3^x)3^x}{9}=1$
$\Rightarrow 3^{2x}-10\cdot 3^x+9=0$
Let $3^x=t$
$\therefore t^2-10t+9=0$
$\Rightarrow t=1$ or $t=9$
$\Rightarrow x=0$ or $x=2$
Hence, the sum of possible values of $x$ is $2$