Let the sixth term in binomial expansion of ${(\sqrt{2^{\log_{2} (10 - 3^{x})}} + \sqrt[5]{2^{(x - 2) \log_{2} 3}})}^{m}$ , in the increasing powers of $2^{(x - 2) \log_{2} 3}$ . 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 _____

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answer is 4.

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Detailed Solution

$\begin{array}{l} 2 {mc}_{2} = {mc}_{1} + {mc}_{3} \\ \frac{2 m (m - 1)}{2} = m + \frac{m (m - 1) (m - 2)}{6} \\ m (m - 1) = m (1 + \frac{(m - 1) (m - 2)}{6}) \\ m - 1 = \frac{6 + (m - 1) (m - 2)}{6} \\ 6 m - 6 = 6 + m^{2} - 3 m + 2 \\ m^{2} - 9 m + 14 = 0 \\ m^{2} - 7 m - 2 m + 14 = 0 \\ (m - 7) (m - 2) = 0 \\ m = 2, 7 \\ m = 2 rejected \\ m = 7 \\ T_{6} =^{7} c_{5} {(\sqrt{2^{\log (10 - 3^{x})}})}^{7 - 5} \times {(\sqrt[5]{2^{(x - 2) \log 3}})}^{5} = 21 \\ 21 = \frac{7 \times 6}{2} (2^{\log_{2} (10 - 3^{x})} \cdot 2^{(x - 2) \log_{2} 3}) \\ 2^{\log (10 - 3^{x}) + (x - 2) \log 3} = 1 \\ \Rightarrow \log (10 - 3^{x}) (3^{x - 2}) = 0 \\ \Rightarrow (10 - 3^{x}) 3^{x} = 9 \\ \Rightarrow (3^{x} - 1) (3^{x} - 9) = 0 \\ 3^{x} = 1, 3^{x} = 3^{2} \Rightarrow x = 0, 2 \\ Sum = 0 + 2^{2} = 4 \end{array}$