Let the sixth term in the binomial expansion | vedclass

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(B) The expansion is $\left((10-3^x)^{1/2} + (3^{x-2})^{1/5}\right)^m$. The general term is $T_{r+1} = {^mC_r} (10-3^x)^{(m-r)/2} (3^{x-2})^{r/5}$.Giv

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$4$

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(B) The expansion is $\left((10-3^x)^{1/2} + (3^{x-2})^{1/5}\right)^m$. The general term is $T_{r+1} = {^mC_r} (10-3^x)^{(m-r)/2} (3^{x-2})^{r/5}$.Given $T_6 = 21$,so $r=5$: ${^mC_5} (10-3^x)^{(m-5)/2} (3^{x-2}) = 21$. Given ${^mC_1}, {^mC_2}, {^mC_3}$ are in $A.P.$,so $2({^mC_2}) = {^mC_1} + {^mC_3}$.$2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6} \implies m^2-m = m + \frac{m^3-3m^2+2m}{6}$.$6m^2-6m = 6m + m^3-3m^2+2m \implies m^3-9m^2+14m = 0$. Since $m \ge 5$,$m^2-9m+14=0 \implies (m-7)(m-2)=0$,so $m=7$.Substituting $m=7$ into $T_6$: ${^7C_5} (10-3^x)^{(7-5)/2} (3^{x-2}) = 21 \implies 21 (10-3^x) \cdot \frac{3^x}{9} = 21$.$(10-3^x) \cdot 3^x = 9 \implies 10 \cdot 3^x - (3^x)^2 = 9 \implies (3^x)^2 - 10 \cdot 3^x + 9 = 0$.$(3^x-9)(3^x-1) = 0 \implies 3^x=9$ or $3^x=1 \implies x=2$ or $x=0$.The sum of the squares of the values is $0^2 + 2^2 = 4$.

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