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(C) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are $^nC_{r-1} 2^{r-1}, ^nC_r 2^r, ^nC_{r+1} 2^{r+1}$.Given the rat

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$1120$

Vedclass · Answer

(C) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are $^nC_{r-1} 2^{r-1}, ^nC_r 2^r, ^nC_{r+1} 2^{r+1}$.Given the ratio $^nC_{r-1} 2^{r-1} : ^nC_r 2^r : ^nC_{r+1} 2^{r+1} = 2 : 5 : 8$.From $\frac{^nC_{r-1} 2^{r-1}}{^nC_r 2^r} = \frac{2}{5}$,we get $\frac{r}{n-r+1} 	imes \frac{1}{2} = \frac{2}{5}$ $\Rightarrow \frac{r}{n-r+1} = \frac{4}{5}$ $\Rightarrow 5r = 4n - 4r + 4$ $\Rightarrow 9r - 4n = 4$ (Eq. $1$).From $\frac{^nC_r 2^r}{^nC_{r+1} 2^{r+1}} = \frac{5}{8}$,we get $\frac{r+1}{n-r} 	imes \frac{1}{2} = \frac{5}{8}$ $\Rightarrow \frac{r+1}{n-r} = \frac{5}{4}$ $\Rightarrow 4r + 4 = 5n - 5r$ $\Rightarrow 9r - 5n = -4$ (Eq. $2$).Subtracting Eq. $2$ from Eq. $1$: $(9r - 4n) - (9r - 5n) = 4 - (-4) \Rightarrow n = 8$.Substituting $n=8$ in Eq. $1$: $9r - 32 = 4$ $\Rightarrow 9r = 36$ $\Rightarrow r = 4$.The middle term coefficient is $^nC_r 2^r = ^8C_4 2^4 = 70 	imes 16 = 1120$.

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term,which is in the middle of these three terms,is $...........$.

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