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Let the sixth term in the binomial expansion of $\left(\sqrt{2^{\log _2}\left(10-3^x\right)}+\sqrt[5]{2^{(x-2) \log _2 3}}\right)^m$, in the increasing powers of $2^{(x-2) \log _2 3}$, be $21$ . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an $A.P.$, then the sum of the squares of all possible values of $x$ is $.........$.
$6$
$4$
$8$
$2$
Solution
$T _6={ }^{ m } C _{ o }\left(10-3^{ x }\right)^{\frac{ m -5}{2}} \cdot\left(3^{ x -2}\right)=21$
${ }^{ m } C _1,{ }^{ m } C _2,{ }^{ m } C _3$ are in $A.P.$
$2.$ ${ }^{ m } C _2={ }^{ m } C _1+{ }^{ m } C _3$
Solving for $m$, we get
$m =$ $2$(rejected), $7$
Put in equation $(1)$
$21 .\left(10-3^x\right) \frac{3^x}{9}=21$
$3^x=3^0, 3^2$
$x =0,2$
