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(C) Let the three consecutive coefficients be $^nC_{r-1} = 28, ^nC_r = 56,$ and $^nC_{r+1} = 70.$Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n

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$8$

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(C) Let the three consecutive coefficients be $^nC_{r-1} = 28, ^nC_r = 56,$ and $^nC_{r+1} = 70.$Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r},$ we have:$\frac{56}{28} = \frac{n-r+1}{r}$ $\Rightarrow 2 = \frac{n-r+1}{r}$ $\Rightarrow 2r = n-r+1$ $\Rightarrow n = 3r - 1.$Using the property $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1},$ we have:$\frac{70}{56} = \frac{n-r}{r+1}$ $\Rightarrow \frac{5}{4} = \frac{n-r}{r+1}$ $\Rightarrow 5r + 5 = 4n - 4r$ $\Rightarrow 4n = 9r + 5.$Substituting $n = 3r - 1$ into the second equation:$4(3r - 1) = 9r + 5$ $\Rightarrow 12r - 4 = 9r + 5$ $\Rightarrow 3r = 9$ $\Rightarrow r = 3.$Now,find $n$:$n = 3(3) - 1 = 8.$

If the three consecutive coefficients in the expansion of $(1 + x)^n$ are $28, 56$ and $70,$ then the value of $n$ is

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