Let the sixth term in the binomial expansion of (√2^ log_2(10-3^x)+√[5]2^(x-2) log_23)^m powers of 2^(x-2) log_23, be 21 . If the binomial…

Given,

Binomial expression,

$${(\sqrt{{2}^{{\mathrm{log}}_{2}(10-{3}^{x})}}+\sqrt[5]{{2}^{(x-2){\mathrm{log}}_{2}3}})}^{m}$$(2log2​(10−3x)​+52(x−2)log2​3​)m

$${(\sqrt{(10-{3}^{x})}+\sqrt[5]{{3}^{(x-2)}})}^{m}$$((10−3x)​+53(x−2)​)m

Now, ${T}_{6}=C5m{(10-{3}^{x})}^{\frac{m-5}{2}}\cdot ({3}^{x-2})=21\ldots (1)$T6​=C5m(10−3x)2m−5​⋅(3x−2)=21…(1)

Also given,

$C1m,C2m,C3m$C1m,C2m,C3m are in A.P.

So, $2\cdot C2m=C1m+C3m$2⋅C2m=C1m+C3m

$$\Rightarrow 2\times \frac{m!}{2!(m-2)!}=m+\frac{m!}{3!(m-3)!}$$⇒2×2!(m−2)!m!​=m+3!(m−3)!m!​

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

Put in equation $(1)$(1)

$$21\cdot (10-{3}^{x})\frac{{3}^{x}}{9}=21$$21⋅(10−3x)93x​=21

$$\Rightarrow (10-{3}^{x}){3}^{x}=9\times 1$$⇒(10−3x)3x=9×1

$$\Rightarrow {3}^{x}={3}^{0},{3}^{2}$$⇒3x=30,32

$$\Rightarrow x=0,2$$⇒x=0,2

Sum of the squares of all possible values of $x=4$x=4.