Question

$9^{th}$ term in the expansion of ${(\frac{x}{a}-\frac{3a}{x^2})}^{12}$ is:

• A. ${}^{12}{C}_9\cdot 3^9{x}^{-12}{a}^6$
• B. ${}^{12}{C}_6\cdot 3^8{x}^{-16}{a}^6$
• C. ${}^{12}{C}_4\cdot 3^8{x}^{-12}{a}^4$
• D. none of the above

Answer 1

The correct option is C ${}^{12}{C}_4\cdot 3^8{x}^{-12}{a}^4$

Given

${(\frac{x}{a}-\frac{3a}{x^2})}^{12}$

General term of given expansion is

$T_{r+1}={\sum }_{r=0}^{12}{}^{12}{C}_r{\left(\frac{x}{a}\right)}^{12-r}{\left(\frac{3a}{x^2}\right)}^r(-1{)}^r$

We need $9^{th}$ term hence put $r=8$

$T_{8+1}={}^{12}{C}_8{\left(\frac{x}{a}\right)}^{12-8}{\left(\frac{3a}{x^2}\right)}^8(-1{)}^8$

$T_9={}^{12}{C}_8{\left(\frac{x}{a}\right)}^4{\left(\frac{3a}{x^2}\right)}^8$

$T_9={}^{12}{C}_8{x}^{4-16}\cdot 3^8\cdot a^{8-4}$

$T_9={}^{12}{C}_8\cdot 3^8{x}^{-12}\cdot a^4$

But we know that ${}^n{C}_r={}^n{C}_{n-r}$

${}^{12}{C}_8={}^n{C}_4$

So

$T_9={}^{12}{C}_4\cdot 3^8{x}^{-12}\cdot a^4$