If the coefficients of x^5 and x^6 are equal | vedclass

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(D) The general term in the expansion of $(a+\frac{x}{5})^{65}$ is $T_{r+1} = {}^{65}C_r a^{65-r} (\frac{x}{5})^r = {}^{65}C_r a^{65-r} \frac{x^r}{5^r

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$\frac{24}{25}$

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(D) The general term in the expansion of $(a+\frac{x}{5})^{65}$ is $T_{r+1} = {}^{65}C_r a^{65-r} (\frac{x}{5})^r = {}^{65}C_r a^{65-r} \frac{x^r}{5^r}$.Coefficient of $x^5$ is ${}^{65}C_5 a^{60} \frac{1}{5^5}$.Coefficient of $x^6$ is ${}^{65}C_6 a^{59} \frac{1}{5^6}$.Given that these coefficients are equal:${}^{65}C_5 \frac{a^{60}}{5^5} = {}^{65}C_6 \frac{a^{59}}{5^6}$.$a = \frac{{}^{65}C_6}{{}^{65}C_5} 	imes \frac{5^5}{5^6} = \frac{65-5}{6} 	imes \frac{1}{5} = \frac{60}{6 	imes 5} = 2$.Now,we need the coefficient of $x^2$ in the expansion of $(2+\frac{x}{5})^4$.The general term is $T_{r+1} = {}^{4}C_r (2)^{4-r} (\frac{x}{5})^r$.For $x^2$,we set $r=2$:Coefficient $= {}^{4}C_2 (2)^{4-2} (\frac{1}{5})^2 = 6 	imes 2^2 	imes \frac{1}{25} = 6 	imes 4 	imes \frac{1}{25} = \frac{24}{25}$.

If the coefficients of $x^5$ and $x^6$ are equal in the expansion of $(a+\frac{x}{5})^{65}$,then the coefficient of $x^2$ in the expansion of $(a+\frac{x}{5})^4$ is

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