Q58 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

If $5|b| < 2|a|$. then the 4 th term in the expansion of $(2 a+5 b)^{-4}$ is

  • A
    ${ }^4 C_3 2^5 5^3 a^5 b^3$
  • Correct answer
    B
    $-{ }^6 C_3 \frac{5^3}{2^7} \frac{b^3}{a^7}$
  • C
    $-{ }^6 C_3 \frac{5^4}{2^8} \frac{b^4}{a^8}$
  • D
    ${ }^4 C_3 2^4 5^4 a^4 b^4$

Solution

Given, binomial can be written as $(2 a)^{-4}\left(1+\frac{5 b}{2 a}\right)^{-4}$, where $\left|\frac{5 b}{2 a}\right| < 1$ So, 4 th term will be $(2 a)^{-4}\left[\frac{(-4)(-5)(-6)}{3 !}\left(\frac{5 b}{2 a}\right)^3\right]$ $\left(\begin{array}{l}\because(1+x)^{-n} \\ =1+n x+\frac{n(n-1)}{2 !} x^2+ \\ \frac{n(n-1)(n-2)}{3 !} x^3+\ldots \infty \\ \text { Where, }|x| < 1\end{array}\right)$ $=\frac{1}{2^4 a^4}\left[(-1) \times \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 ! \times 3 \times 2 \times 1} \times \frac{5^3 b^3}{2^3 \cdot a^3}\right]$ $=-{ }^6 C_3 \cdot \frac{5^3 b^3}{2^7 \cdot a^7}$

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

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