The numerically greatest term in the expansion ${(2 x - 3 y)}^{12}$ when $x = 1$ and $y = 5 / 2$ is the

11

th term

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10

th term

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9

th term

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6

th term

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Solution

The correct option is A $11$ th term
General term in expansion of $(a + b)^{n}$ is given
by $T_{r + 1} =^{n} C_{r} a^{r} . b^{n - r}$
For numerically greatest term, finding
$∣ ∣ \frac{T_{r + 1}}{T_{r}} ∣ ∣ = ∣ ∣ ∣ \frac{^{n} C_{r} . a^{r} . b^{n - r}}{^{n} C_{r - 1} . a^{r - 1} b^{n - r + 1}} ∣ ∣ ∣ = ∣ ∣ \frac{^{n} C_{r}}{^{n} C_{r - 1}} ∣ ∣ ∣ ∣ \frac{a}{b} ∣ ∣ = ∣ ∣ \frac{n - r + 1}{r} ∣ ∣ ∣ ∣ \frac{a}{b} ∣ ∣$
Here, $∣ ∣ \frac{T_{r + 1}}{T_{r}} ∣ ∣ = ∣ ∣ \frac{12 - r + 1}{r} ∣ ∣ ∣ ∣ ∣ \frac{2 (1)}{- 3 (\frac{5}{2})} ∣ ∣ ∣ = ∣ ∣ \frac{13 - r}{r} ∣ ∣ ∣ ∣ \frac{4}{15} ∣ ∣ > 1$
$(13 - r) 4 > 15 r \Rightarrow 52 - 4 r > 15 r \Rightarrow 52 / 19 \Rightarrow r = 2$
$\Rightarrow T_{3}$ or $T_{11}$ term is greatest (In expansion
terms of begining are equal to that from last)

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