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
B
n=2
C
n=7
D
n=5

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 $$\Rightarrow \dfrac{9(n-1)}{4}=3(n-2)$$$$\Rightarrow 3n-3=4n-8$$ $$\Rightarrow n=5$$ Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More