# Question 5 Exercise 7.2

Solutions of Question 5 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 5(i)

Find middle term in the expansion of $(\dfrac{a}{x}+b x)^8$

### Solution

Since we see that $a=\dfrac{a}{x}$. $b=b x$ and $n=8$

Since $n-8$ is a the even number of terms in the expansion are $8+1=9$ The middle term is only one so, $$(\dfrac{8+2}{2})^{t h}=5^{t h}$.

Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{8 !}{(8-r) ! r !}(\dfrac{a}{x})^{8-r}(b x)^r$$ To get middle term $T_5$, we put $r=4$ \begin{align}T_5&=\dfrac{8 !}{(8-4) ! 4 !}(\dfrac{a}{x})^{8-4}(b x)^4 \\ & =70 \cdot \dfrac{a^4}{x^4} \cdot b^4 x^4 \\ & =70 \cdot a^4 b^4 \end{align} Thus $T_5$ is the middle term of the expansion which is $70 a^4 b^4$.

## Question 5(ii)

Find middle term in the expansion of $(3 x-\dfrac{x^2}{2})^9$

### Solution

Since we see that $a=3 x$, $b=-\dfrac{x^2}{2}$ and $n=9$.

Since $n=9$ is odd so the total number of terms in the expansion are $9+1=10$.

So in this we have two middle terms that are $(\dfrac{9+1}{2})^{t h}=5^{\text {th }}$ and $(\dfrac{9+3}{2})^{t h}=6^{t h}$

The general term is: $$T_{r+1}=\dfrac{9 !}{(9-r) ! r !}(3 x)^{9-r}(-\dfrac{x^2}{2})^r$$ Putting $r=4$ to get first middle term \begin{align} T_5&=\dfrac{9 !}{(9-4) ! 4 !}(3 x)^{9-4}(-\dfrac{x^2}{2})^4 \\ & =126 \cdot 3^5 \cdot x^5 \cdot(-1)^4 \cdot \dfrac{(x^2)^4}{2^4} \\ & T_5=\dfrac{30618}{16} x^{13}\\ & T_5=\dfrac{15309}{8} x^{13}\end{align} Putting $r=5$ to get the $2^{\text {nd }}$ middle term that is: \begin{align}T_6&=\dfrac{9 !}{(9-5) ! 5 !}(3 x)^{9-5}(-\dfrac{x^2}{2})^5 \\ & =126 \cdot(-1)^5 \cdot 3^4 \cdot x^4 \cdot \dfrac{1}{2^5} \cdot(x^2)^5\\ &=-\dfrac{5103}{16} x^{14} \end{align} Thus the two middle terms are $$T_5=\dfrac{15309}{8} x^{13} \text { and } T_6=-\dfrac{5103}{16} x^{14}$$

## Question 5(iii)

Find middle term in the expansion of $(3 x^2-\dfrac{y}{3})^{10}$

### Solution

Since we see that $a=3 x^2$, $b=-\dfrac{y}{3}$ and $n=10$.

Since $n-10$ is even so the total number of terms in the expansion are $10_{\neg} 1=11$.

The middle term is only one and is $(\dfrac{10+2}{2})^{t h}=6^{t h}$.

Now $T_{r+1}$ of the given expansion is: $$T_{r+1}=\dfrac{10 !}{(10-r) ! r !}(3 x^2)^{10-r}(-\dfrac{y}{3})^r$$ To get middle term, putting $r=5$ \begin{align}T_6&=\dfrac{10 !}{(10-5) ! 5 !}(3 x^2)^{10-5}(-\dfrac{y}{3})^5 \\ T_6&=-252.3^5 \cdot \dfrac{1}{3^5} \cdot(x^2)^5 \cdot y^5 \\ \rightarrow T_6&=-252 x^{10} y^5 \end{align} is the required middle term.

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