If the $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms in the expansion of…

If the $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms in the expansion of $(x+a)^{\mathrm{n}}$ are $96,216,216$ respectively and $n$ is a positive integer then $a+x=$
  1. $n+1$
  2. $n$
  3. $n-1$
  4. $\frac{n}{2}$

Solution

$\begin{aligned} & { }^n C_1 x^{n-1} a=96 ....(i)\\ & { }^n C_2 x^{n-2} a^2=216 ....(ii)\\ & { }^n C_3 x^{n-3} a^3=216 .....(iii) \end{aligned}$ (i) $\div$ (ii) we get, $\begin{aligned} & \frac{{ }^n C_1 x}{{ }^n C_2 a}=\frac{96}{216} \Rightarrow \frac{2 n}{n(n-1)} \frac{x}{a}=\frac{16}{36}=\frac{4}{9} \\ & \Rightarrow 9 x=2(n-1) a \end{aligned}$ (ii) $\div$ (iii) we get, $\begin{aligned} & \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{a}=1 \Rightarrow \frac{n(n-1)}{2} x=\frac{n(n-1)(n-2) a}{6} \\ & \Rightarrow 3 x=a(n-2) \\ & \frac{9 x}{3 x}=\frac{2(n-1) a}{a(n-2)} \Rightarrow 3=\frac{2(n-1)}{(n-2)} \\ & \Rightarrow 3 n-6=2 n-2 \Rightarrow n=4 . \\ & \frac{2 x}{a(n-1)}=\frac{4}{9} \\ & \Rightarrow 9 x=2 a \times 3=6 a \Rightarrow 3 x=2 a \\ & { }^n C_1 x^{n-1} a=96 \Rightarrow 4 \times x^3 \times \frac{3}{2} x=96 \\ & \Rightarrow x^4=16 \Rightarrow x=2 \Rightarrow a=3 \\ & \Rightarrow a+x=5=n+1 . \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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