Let the sixth term in the binomial expansion of ( 2 ^ _ 2 (10 - 3 ^...

Given,

Binomial expression,

${\left(\sqrt{{2}^{\log_{2}\left(10 - {3}^{x}\right)}} + {2}^{( x - 2 ) \log_{2}3}5\right)}^{m}$

${\left(\sqrt{\left(10 - {3}^{x}\right)} + {3}^{( x - 2 )}5\right)}^{m}$

Now, ${T}_{6} = C5 m \left(10 - {3}^{x}\right)\dfrac{m - 5}{2} \cdot \left({3}^{x - 2}\right) = 21 \ldots \left(1\right)$

Also given,

$C1 m , C2 m , C3 m$ are in A.P.

So, $2 \cdot C2 m = C1 m + C3 m$

$⇒ 2 \times \dfrac{m !}{2 ! \left(m - 2\right) !} = m + \dfrac{m !}{3 ! \left(m - 3\right) !}$

Solving for $m$, we get $m = 2, 7 and m = 2$ (rejected), so $m = 7$

Put in equation \left(1\right)$</p><p>$21 \cdot \left(10 - {3}^{x}\right)\dfrac{{3}^{x}}{9} = 21$</p><p>$⇒ \left(10 - {3}^{x}\right){3}^{x} = 9 \times 1$</p><p>$⇒ {3}^{x} = {3}^{0}, {3}^{2}$</p><p>$⇒ x = 0, 2$</p><p>Sum of the squares of all possible values of$x = 4$.