Verified Solution • 2025–26 Session moderate

QJEEM05Madcc6: JEE Main 2005 Mathematics - Class 11 JEEM05Madcc6 Solution: If the coefficients of r th , (r+1) th , and (r + 2) th terms…

If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of (1+y)m{{\rm{(1 + y )}}^m} are in A.P., then m and r satisfy the equation

Multiple Choice Options

  • A m2−m(4r−1)+4 r2−2=0{m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0
  • B m2−m(4r+1)+4 r2+2=0{m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0
  • C m2−m(4r+1)+4 r2−2=0{m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0
  • D m2−m(4r−1)+4 r2+2=0{m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0

Step-by-Step Verified Solution

Let r = 2

∴\therefore 2nd, 3rd and 4th terms are in AP.

2nd term = T2 = mC1.y{}^m{C_1}.y

Coefficient of T2 = mC1{}^m{C_1}

3rd term = T3 = mC2.y2{}^m{C_2}.{y^2}

Coefficient of T3 = mC2{}^m{C_2}

4th term = T4 = mC3.y3{}^m{C_3}.{y^3}

Coefficient of T2 = mC3{}^m{C_3}

∴\therefore 2. mC2{}^m{C_2} = mC1{}^m{C_1} + mC3{}^m{C_3}

⇒\Rightarrow 2.m(m−1)1.22.{{m\left( {m - 1} \right)} \over {1.2}} = m1{m \over 1} + m(m−1)(m−2)1.2.3{{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}

⇒\Rightarrow 6m2 - 6m = 6m +m(m2 - 3m + 2)

⇒\Rightarrow 6m2 - 6m = 6m + m3 - 3m2 + 2m

⇒\Rightarrow 6m - 6 = 6 + m2 - 3m + 2

⇒\Rightarrow m2 - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, m2−m(4r+1)+4 r2−2=0{m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0 putting r = 2 we get

m2 - 9m + 14 = 0. So Option C is correct.

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