JEE Main 2021 — Mathematics Algebra

The general term in the expansion of ${(a+b)}^{n}$(a+b)n is ${T}_{r+1}=Crn{a}^{n-r}{b}^{r}.$Tr+1​=Crnan−rbr.

Hence, the general term in the expansion of ${(x+\frac{a}{{x}^{2}})}^{n}$(x+x2a​)n is ${T}_{r+1}=Crn(x{)}^{n-r}{(\frac{a}{{x}^{2}})}^{r}$Tr+1​=Crn(x)n−r(x2a​)r

$$\Rightarrow {T}_{r+1}=Crn{a}^{r}{x}^{n-3r}$$⇒Tr+1​=Crnarxn−3r

Given the coefficients of third, fourth and fifth terms are in the ratio $12:8:3$12:8:3

$$\Rightarrow (C2n{a}^{2}):(C3n{a}^{3}):C4n{a}^{4}=12:8:3$$⇒(C2na2):(C3na3):C4na4=12:8:3

$\Rightarrow \frac{C2n{a}^{2}}{C3n{a}^{3}}=\frac{12}{8}$⇒C3na3C2na2​=812​ and $\frac{C3n{a}^{3}}{C4n{a}^{4}}=\frac{8}{3}$C4na4C3na3​=38​

Using $\frac{Crn}{Cr+1n}=\frac{r+1}{n-r},$Cr+1nCrn​=n−rr+1​,

$\Rightarrow \frac{3}{(n-2)a}=\frac{3}{2}$⇒(n−2)a3​=23​ and $\frac{4}{(n-3)a}=\frac{8}{3}$(n−3)a4​=38​

On dividing the two equations, we get $\frac{3(n-3)}{4(n-2)}=\frac{9}{16}$4(n−2)3(n−3)​=169​

$$\Rightarrow 4n-12=3n-6$$⇒4n−12=3n−6

$\Rightarrow n=6$⇒n=6 and $a=\frac{1}{2}.$a=21​.

For term independent of $x\Rightarrow n=3r$x⇒n=3r

$$\Rightarrow r=2.$$⇒r=2.

$\therefore$∴ The coefficient is$C26{(\frac{1}{2})}^{2}=\frac{15}{4}.$C26(21​)2=415​.

Hence, the nearest integer is $4.$4.