If the 9^{th} and 10^{th} terms are the numer | vedclass

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(B) For the expansion $(ax + by)^n$,the $r^{th}$ and $(r+1)^{th}$ terms are numerically greatest if $r = \frac{(n+1) \cdot |by/ax|}{1 + |by/ax|}$.Give

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$^{14}C_7 6^7$

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(B) For the expansion $(ax + by)^n$,the $r^{th}$ and $(r+1)^{th}$ terms are numerically greatest if $r = \frac{(n+1) \cdot |by/ax|}{1 + |by/ax|}$.Given $x = 2/5$ and $y = 1/2$,we have $|by/ax| = |(6 	imes 1/2) / (5 	imes 2/5)| = 3/2$.Since the $9^{th}$ and $10^{th}$ terms are greatest,$r = 9$.$9 = \frac{(n+1)(3/2)}{1 + 3/2} = \frac{(n+1)(3/2)}{5/2} = \frac{3(n+1)}{5}$.$45 = 3(n+1)$ $\Rightarrow n+1 = 15$ $\Rightarrow n = 14$.The middle term of $(5x - 6y)^{14}$ is the $(14/2 + 1)^{th} = 8^{th}$ term.However,the question asks for the absolute value of the middle term,which is $|T_8| = |^{14}C_7 (5x)^7 (-6y)^7|$.$|T_8| = ^{14}C_7 (5 	imes 2/5)^7 (6 	imes 1/2)^7 = ^{14}C_7 (2)^7 (3)^7 = ^{14}C_7 6^7$.

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

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