Let the coefficients of the middle terms in the expansion of \(\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}\) and \(\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0\), respectively form the first three terms of an A.P. If d is the common difference of this A.P., then \(50-\frac{2 \mathrm{~d}}{\beta^{2}}\) is equal to

Binomial Theorem — JEE Main · 2022 · 28 Jul 2022 · Shift 2 Question with Solution

Let the coefficients of the middle terms in the expansion of

(\frac{1}{6} + β x)^{4}, (1 - 3 β x)^{2}

and

(1 - \frac{β}{2} x)^{6}, β > 0

, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then

50 - \frac{2 d}{β ^{2}}

is equal to

Answer: 57

The middle term coefficients are:

(2 4) \cdot \frac{β ^{2}}{6} = \frac{β ^{2}}{1}

- 6 β

(coefficient of

x

(1 - 3 β x)^{2}

), and

- (3 6) \cdot \frac{β ^{3}}{8} = - \frac{5 β ^{3}}{2}

. For these to be in A.P.:

2 (- 6 β) = \frac{β ^{2}}{1} - \frac{5 β ^{3}}{2}

. Simplifying:

β^{2} - \frac{5}{2} β^{3} = - 12 β

. Since

β > 0

, dividing by

β

β - \frac{5}{2} β^{2} = - 12

, which gives

5 β^{2} - 2 β - 24 = 0

. Solving:

β = \frac{12}{5}

(taking the positive root). The common difference is

d = - 6 β - \frac{β ^{2}}{1} = - \frac{72}{5} - \frac{144}{25} = - \frac{504}{25}

. Therefore,

50 - \frac{2 d}{β ^{2}} = 50 - \frac{2 \cdot ( - 504/25 )}{( 12/5 ) ^{2}} = 50 + \frac{1008/25}{144/25} = 50 + 7 = 57

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer