The numerically greatest term in the expansion ${(5 x - 6 y)}^{14}$ when $x = \frac{2}{5}, y = \frac{1}{2}$ is

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$^{14} C_{7} 2^{6} 3^{8}$

$^{14} C_{6} 2^{8} 3^{6}$

$^{14} C_{7} 2^{8} 3^{6}$

$^{14} C_{8} 2^{6} 3^{8}$

answer is C.

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Detailed Solution

Given expansion ${(5 x - 6 y)}^{14}$ is

${(5 x - 6 y)}^{14} = {(5 x)}^{14} {(1 - \frac{6 y}{5 x})}^{14}$ this expansion covert into ${(1 + x)}^{n}$

The numerically greatest term in the expansion of ${(1 + x)}^{n}$ is $\frac{(n + 1) | x |}{| x | + 1}$

$N . G . T = \frac{15 | \frac{6 \times \frac{1}{2}}{5 \times \frac{2}{5}} |}{\frac{3}{2} + 1} = 9$

We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

$T_{9} = T_{8 + 1} =^{14} C_{8} {(5 x)}^{6} {(- 6 y)}^{8} ........................ (1)$

Substitute these values in Eqn (1) $x = \frac{2}{5}, y = \frac{1}{2}$ then we get

$T_{9} =^{14} C_{8} 2^{6} \times 3^{8}$ is the numerically greatest term in the given expansion

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