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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2024 · Mathematics · Binomial Theorem

If the 4th term in the expansion of (m x+(1)/(x))^n is (5)/(2) ,...

If the 4th term in the expansion of (mx+1x)n\left(m x+\frac{1}{x}\right)^n is 52\frac{5}{2}, then that is the value of mnm n ?

Options

  1. A-3
  2. B3
  3. C6
  4. D12

Correct answer

B. 3

Step-by-step solution

T4=T3+1=nC3(mx)n−3(1x)3=5252x0=nC3(m)n−3xn−6\begin{aligned} \mathrm{T}_4 & =\mathrm{T}_3+1 \\ & ={ }^n \mathrm{C}_3(m x)^{n-3}\left(\frac{1}{x}\right)^3=\frac{5}{2} \\ \frac{5}{2} x^0 & ={ }^n \mathrm{C}_3(m)^{n-3} x^{n-6}\end{aligned} On comparing the power of xx∴n−6=0⇒n=6 and 6C3m3=52⇒6.5.43.2m3=5220m2=52⇒m3=18⇒m=12∴mn=12×6=3. \begin{aligned} & \therefore \quad n-6=0 \\ & \Rightarrow \quad n=6 \\ & \text{ and } \quad{ }^6 \mathrm{C}_3 m^3=\frac{5}{2} \\ & \Rightarrow \quad \frac{6.5.4}{3.2} m^3=\frac{5}{2} \\ & 20 m^2=\frac{5}{2} \\ & \Rightarrow \quad m^3=\frac{1}{8} \\ & \Rightarrow \quad m=\frac{1}{2} \\ & \therefore \quad m n=\frac{1}{2} \times 6=3 \text{. } \end{aligned}$$
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