QJEEM21Ma250a: JEE Main 2021 Mathematics - Class 11 JEEM21Ma250a Solution: Let the coefficients of third, fourth and fifth terms in the…

Question

Verified Solution • 2025–26 Session moderate

Let the coefficients of third, fourth and fifth terms in the expansion of

{\left( {x + {a \over {{x^2}}}} \right)^n},x \ne 0

, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________.

Accepted Answer

{T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\left( {{a \over {{x^2}}}} \right)^r}

= {}^n{C_r}{a^r}{x^{n - 3r}}

{T_3} = {}^n{C_2}{a^2}{x^{n - 6}}

,

{T_4} = {}^n{C_3}{a^3}{x^{n - 9}}

,

{T_5} = {}^n{C_4}{a^4}{x^{n - 12}}

Now,

{{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}

\Rightarrow a(n - 2) = 2

.......... (i)

and

{{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}

\Rightarrow a(n - 3) = {3 \over 2}

........ (ii)

by (i) and (ii)

n = 6,\,a = {1 \over 2}

for term independent of 'x'

n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2

{T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4

Step-by-Step Verified Solution