Q57 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

If the 9 th 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

  • A
    $14 C_8 6^7$
  • Correct answer
    B
    $14 C_7 6^7$
  • C
    $15 C_7 6^7$
  • D
    $15 C_8 6^8$

Solution

Given, 9th and 10th terms are numerically greatest terms in the expansion of $(5 x-6 y)^n$ $9 \leq \frac{(n+1)\left(\frac{6 y}{5 x}\right)}{1+\left(\frac{6 y}{5 x}\right)}$ Here, $y=\frac{1}{2}$ and $x=\frac{2}{5}$ $\because \quad \frac{y}{x}=\frac{5}{4} \Rightarrow \frac{6 y}{5 x}=\frac{3}{2}$ $9 \leq \frac{(n+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}$ $9 \leq \frac{3(n+1)}{5}$ $n+1 \geq 15$ $n \geq 14$ $\because \quad n=14$ Middle term of $(5 x-6 y)^{14}$ $=\left|{ }^{14} C_7(5 x)^7(-6 y)^7\right|$ $=\left|{ }^{14} C_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} C_7 \times 6^7$

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

Relevant Questions