Q60 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

The numerically greatest term in the expansion of $(3 x-16 y)^{15}$ when $x=\frac{2}{3}$ and $y=\frac{3}{2}$ is

  • A
    13th term
  • B
    14th term
  • Correct answer
    C
    15th term
  • D
    16th term

Solution

$(3 x-16 y)^{15}=(3 x)^{15}\left(1-\frac{16 y}{3 x}\right)^{15}=(3 x)^{15}(1+\alpha)^n$ Here $n=15, \alpha=-\frac{16 y}{3 x}=\frac{-16 \times \frac{3}{2}}{3 \times \frac{2}{3}}=\frac{-24}{2}=-12$ $\frac{(n+1)|\alpha|}{|\alpha|+1}=\frac{(15+1) \cdot 12}{12+1}=14.76$ $\therefore 15^{\text {th }}$ term is numerically greatest.

Hence, the correct answer is option: 'C'.

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