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Question

If the second, third and fourth terms in the expansion of $$(b + a)^n$$ are $$135, 30$$, and $$\dfrac{10}{3}$$ respectively, then:


A
n=3
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B
n=2
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C
n=7
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D
n=5
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Solution

The correct option is D $$n=5$$
Given 
$$(b+a)^n=(a+b)^n$$
The second term is 
$$T_{2}=\displaystyle^nC_{1}a^{n-1}b=135$$

$$T_{2}=na^{n-1}b=135$$

The third term is 
$$T_{3}=\displaystyle^nC_{2}a^{n-2}b^2=30$$

$$T_{3}=\dfrac{n(n-1)a^{n-2}b^2}{2}=30$$

The fourth term is 
$$T_{4}=\displaystyle^nC_{3}a^{n-3}b^3=\dfrac{10}{3}$$

$$T_{4}=\dfrac{n(n-1)(n-2)a^{n-3}b^3}{6}=\dfrac{10}{3}$$

On taking ratio of $$T_{2},T_{3}$$
$$\Rightarrow \dfrac{T_{2}}{T_{3}}=\dfrac{135}{30}$$

$$\Rightarrow \dfrac{na^{n-1}b}{\dfrac{n(n-1)a^{n-2}b^2}{2}}=\dfrac{135}{30}$$

$$\Rightarrow \dfrac{2a}{(n-1)b}=\dfrac{9}{2}$$

$$\Rightarrow 4a=9b(n-1)------(1)$$

On taking ratio of $$T_{3},T_{4}$$
$$\Rightarrow \dfrac{T_{3}}{T_{4}}=\dfrac{3(30)}{10}$$

$$\Rightarrow \dfrac{\dfrac{n(n-1)a^{n-2}b^2}{2}}{\dfrac{n(n-1)(n-2)a^{n-3}b^3}{6}}=9$$

$$\Rightarrow \dfrac{3a}{(n-2)b}=9$$

$$\Rightarrow a=3b(n-2)$$

$$\Rightarrow \dfrac{a}{b}=3(n-2)----(2)$$

Substituting $$\dfrac{a}{b}$$ in eq (1), we get 

$$\Rightarrow \dfrac{9(n-1)}{4}=3(n-2)$$

$$\Rightarrow 3n-3=4n-8$$ 

$$\Rightarrow n=5$$ 

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