Question

Let the coefficients of the middle terms in the expansion of ${\displaystyle {(\frac{1}{\sqrt{6}}+\beta x)}^4,{(1-3\beta x)}^2}$ and ${\displaystyle {(1-\frac{\beta }{2}x)}^6,\beta >0}$, common difference of this A.P., then ${\displaystyle 50-\frac{2d}{{\beta }^2}}$ is equal to ______.

Options

• 56

• 57

• 58

• 59

Answer 1

Let the coefficients of the middle terms in the expansion of ${\displaystyle {(\frac{1}{\sqrt{6}}+\beta x)}^4,{(1-3\beta x)}^2}$ and ${\displaystyle {(1-\frac{\beta }{2}x)}^6,\beta >0}$, common difference of this A.P., then ${\displaystyle 50-\frac{2d}{{\beta }^2}}$ is equal to 57.

Explanation:

Coefficient of middle term

${\displaystyle {}^4{C}_2\times \frac{{\beta }^2}{6},-6\beta -{}^6{C}_3\times \frac{{\beta }^3}{8}}$ are in A.P.

2(–6β) = ${\displaystyle {}^4{C}_2\frac{{\beta }^2}{6}-{}^6{C}_3 \frac{{\beta }^3}{8}}$

${\displaystyle {\beta }^2-\frac{5}{2}{\beta }^3}$ = –12β

β = ${\displaystyle \frac{12}{5}}$ or β = –2

∴ β = ${\displaystyle \frac{12}{5}}$

Common difference

d = ${\displaystyle \frac{72}{5}-\frac{144}{25}=-\frac{504}{25}}$

∴  ${\displaystyle 50-\frac{2d}{{\beta }^2}}$ = 57