Question

Let the coefficients of the middle terms in the expansion of `(1/sqrt(6) + βx)^4, (1 - 3βx)^2` and `(1 - β/2x)^6, β > 0`, common difference of this A.P., then `50 - (2d)/β^2` is equal to ______.

विकल्प

• 56

• 57

• 58

• 59

Answer 1

Let the coefficients of the middle terms in the expansion of `(1/sqrt(6) + βx)^4, (1 - 3βx)^2` and `(1 - β/2x)^6, β > 0`, common difference of this A.P., then `50 - (2d)/β^2` is equal to 57.

Explanation:

Coefficient of middle term

`""^4C_2 xx β^2/6, -6β -""^6C_3 xx β^3/8` are in A.P.

2(–6β) = `""^4C_2 β^2/6 - ""^6C_3  β^3/8`

`β^2 - 5/2β^3` = –12β

β = `12/5` or β = –2

∴ β = `12/5`

Common difference

d = `72/5 - 144/25 = -504/25`

∴  `50 - (2d)/β^2` = 57