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(A) Let the three consecutive coefficients be ${}^nC_{r-1}, {}^nC_r, {}^nC_{r+1}$.Given the ratio ${}^nC_{r-1} : {}^nC_r : {}^nC_{r+1} = 1 : 5 : 20$.F

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$3654$

Vedclass · Answer

(A) Let the three consecutive coefficients be ${}^nC_{r-1}, {}^nC_r, {}^nC_{r+1}$.Given the ratio ${}^nC_{r-1} : {}^nC_r : {}^nC_{r+1} = 1 : 5 : 20$.From $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{5}{1}$,we get $\frac{n-r+1}{r} = 5 \implies n-r+1 = 5r \implies n = 6r-1 \dots(1)$.From $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{20}{5} = 4$,we get $\frac{n-r}{r+1} = 4 \implies n-r = 4r+4 \implies n = 5r+4 \dots(2)$.Equating $(1)$ and $(2)$,$6r-1 = 5r+4 \implies r = 5$.Substituting $r=5$ in $(1)$,$n = 6(5)-1 = 29$.The coefficient of the $4^{	ext{th}}$ term is ${}^nC_3 = {}^{29}C_3$.${}^{29}C_3 = \frac{29 	imes 28 	imes 27}{3 	imes 2 	imes 1} = 29 	imes 14 	imes 9 = 3654$.

If the coefficients of the three consecutive terms in the expansion of $(1+x)^n$ are in the ratio $1:5:20$,then the coefficient of the fourth term is $............$.

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