Let {}^nC_{r-1}=28,{}^nC_r=56,and {}^nC_{r+1} | vedclass

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(A) Given ${}^nC_{r-1} = 28$,${}^nC_r = 56$,and ${}^nC_{r+1} = 70$. Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we have: $\frac

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$20$

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(A) Given ${}^nC_{r-1} = 28$,${}^nC_r = 56$,and ${}^nC_{r+1} = 70$. Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we have: $\frac{56}{28} = \frac{n-r+1}{r}$ $\Rightarrow 2r = n-r+1$ $\Rightarrow n = 3r-1$ $(i)$. Similarly,$\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1} = \frac{70}{56} = \frac{5}{4}$. $4(n-r) = 5(r+1)$ $\Rightarrow 4n - 4r = 5r + 5$ $\Rightarrow 4n = 9r + 5$ (ii). Substituting $(i)$ into (ii): $4(3r-1) = 9r+5$ $\Rightarrow 12r - 4 = 9r + 5$ $\Rightarrow 3r = 9$ $\Rightarrow r = 3$. Then $n = 3(3)-1 = 8$. Coordinates of $C$ are $(3(3)-8, 3^2-8-1) = (1, 0)$. The centroid $(x, y)$ of $	riangle ABC$ is given by: $x = \frac{4 \cos t + 2 \sin t + 1}{3} \Rightarrow 3x - 1 = 4 \cos t + 2 \sin t$. $y = \frac{4 \sin t - 2 \cos t + 0}{3} \Rightarrow 3y = 4 \sin t - 2 \cos t$. Squaring and adding: $(3x-1)^2 + (3y)^2 = (4 \cos t + 2 \sin t)^2 + (4 \sin t - 2 \cos t)^2$. $= 16 \cos^2 t + 4 \sin^2 t + 16 \sin^2 t + 4 \cos^2 t = 20(\cos^2 t + \sin^2 t) = 20$. Thus,$\alpha = 20$.

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