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If the $6^{th}$ term in the expansion of the binomial ${\left[ {\sqrt {{2^{\log (10 - {3^x})}}} + \sqrt[5]{{{2^{(x - 2)\log 3}}}}} \right]^m}$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first, third and fifth terms of an $A.P$. (the symbol log stands for logarithm to the base $10$), then $x = $
$0$
$1$
$2$
$a$ or $c$ both
Solution
(d) Since coefficients $^m{C_1},$ $^m{C_2}$and $^m{C_3}$ of ${T_2},{T_3},{T_4}$ i.e.
are the first, third and fifth terms of an $A. P.$, which will also be in $A. P.$ of common difference $2d$.
Hence $2$ $^m{C_2}{ = ^m}{C_1}{ + ^m}{C_3} \Rightarrow (m – 2)(m – 7) = 0$.
Since $6^{th}$ term is $21$, $m = 2$ is ruled out and we have $m = 7$ and ${T_6} = 21{ = ^7}{C_5}{\left[ {\sqrt {{2^{\log (10 – {3^x})}}} } \right]^{7 – 5}} \times {\left[ {\sqrt[5]{{{2^{(x – 2)}}\log 3}}} \right]^5}$
$ \Rightarrow $ $21 = 21.\,{2^{\log (10 – {3^x}) + \log {3^{x – 2}}}}$
==> ${2^{\log [(10 – {3^x})\,\,{3^{x – 2}}]}} = 1 = {2^0}$
Which on simplification gives $x = 0, 2.$
