The greatest term in the expansion of ${(2 x - 3 y)}^{28}$ , when $x = 9$ , $y = 4$ is

14^{t h}

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15^{t h}

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14^{t h}

15^{t h}

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12^{t h}

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Solution

The correct option is D $12^{t h}$
The greatest term will be
$\frac{(n + 1) . | y |}{| x | + | y |}$
$= \frac{(28 + 1) .12}{18 + 12}$
$= \frac{29.12}{30}$
$= \frac{29.4}{10}$
$= 11.6$
Hence the greatest term will be 12th term.

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