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Q.

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

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a

$^{14} C_{_{8}} 6^{7}$

b

$^{14} C_{_{7}} 6^{7}$

c

$^{15} C_{_{7}} 6^{7}$

d

$^{15} C_{_{8}} 6^{8}$

answer is B.

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Detailed Solution

$9 \leq \frac{(n + 1) (\frac{6 y}{5 x})}{1 + (\frac{6 y}{5 x})}$
$\begin{array}{l} H e r e y = \frac{1}{2} a n d x = \frac{2}{5} \\ 9 \leq \frac{(n + 1) (\frac{3}{2})}{1 + \frac{3}{2}} \\ 9 \leq \frac{3 (n + 1)}{5} \end{array}$
$\begin{array}{l} n + 1 \geq 15 \\ n \geq 15 \\ n = 14 \\ m i d d l e t e r m o f {(5x-6y)}^{14} \\ = ||^{14} C_{7} {(5 \times \frac{2}{3})}^{7} {(- 6 \times \frac{1}{8})}^{7}| \\ =^{14} C_{_{7}} \times 6^{7} \end{array}$

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