Question

Find the middle term in the expansion of:

$\left(i\right){(\frac{2}{3}x-\frac{3}{2x})}^{20}$

$\left(ii\right){(\frac{a}{x}+bx)}^{12}$

$\left(iii\right){(x^2-\frac{2}{x})}^{10}$

$\left(iv\right){(\frac{x}{a}-\frac{a}{x})}^{10}$

Answer 1

$\left(i\right){(\frac{2}{3}x-\frac{3}{2x})}^{20}$

Here, n=20 which is an even number so, ${(\frac{20}{2}+1)}^{th}i.e.,{11}^{th}$ term is the middle term We know that,

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^{r}n=20,r=10,x=\frac{2}{3}x,Y=\frac{2}{3x}$

$T_{11}=T_{10+1}=(-1{)}^{10}{}^{20}{C}_{10}{\left(\frac{2}{3}x\right)}^{10}{\left(\frac{3}{2x}\right)}^{10}$

$={}^{20}{C}_{10}\frac{2^{10}}{3^{10}}\times \frac{3^{10}}{2^{10}}\times \frac{x^{10}}{x^{10}}={}^{20}{C}_{10}$

$\left(ii\right){(\frac{a}{x}+bx)}^{12}$

Here, n =12, which even number

So,${(\frac{12}{2}+1)}^{th}$ term i.e., $7^{th}$ term is the middle term.

Hence, the middle term $=T_7=T_{6+1}$

$\therefore T_7=T_{6+1}={}^{12}{C}_6\times {\left(\frac{a}{x}\right)}^{12-6}\times (bx{)}^6$

$={}^{12}{C}_6{\left(\frac{a}{x}\right)}^6\times (bx{)}^6$

$=\frac{12!}{(12-6)!6!}\times \frac{a^6}{x^6}\times b^6\times 6$

$=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{(6\times 5\times 4\times 3\times 2\times 1)}\times a^6{b}^6$

$=924\times a^6{b}^6$

$\therefore$ The middle term $=924\times a^6{b}^6.$

$\left(iii\right){(x^2-\frac{2}{x})}^{10}Here,n=10$

$\therefore {(\frac{n}{2}+1)}^{th}={(\frac{10}{2}+1)}^{th}=6^{th}$ term is the middle term.

The term formula is

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_5{x}^{n-r}{y}^r$

$T_6=T_{5+1}=(-1{)}^5{}^{10}{C}_5(x^2{)}^{10-5}{\left(\frac{2}{x}\right)}^5$

$={-}^{10}{C}_5{x}^{20-10}\frac{2^5}{x^5}$

$=\frac{-10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times 2^5{x}^5=-8064x^5$

$\left(iv\right){(\frac{x}{a}-\frac{a}{x})}^{10}$

Here n =10, which is even, therefore it has 11 terms

$\therefore middletermis(\frac{n}{2}+1)=6^A$ term

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_6=T_{5+1}=(-1{)}^5{}^{10}{C}_5{\left(\frac{x}{a}\right)}^{10-5}{\left(\frac{x}{a}\right)}^5$

$=-\frac{10!}{5!5!}\times \frac{x^5}{a^3}\times a^5\times x^{-5}$

=-252