Question

The positive value of ${}^{\prime }{a}^{\prime }$ so that the coefficient of $x^5$ is equal to that of $x^{15}$ in the expansion of $(x^2+\frac{a}{x^3}{)}^{10}$ is

• A. $\frac{1}{2\sqrt{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $1$
• D. $2\sqrt{3}$

Answer 1

The correct option is A $\frac{1}{2\sqrt{3}}$

${(x^2+\frac{a}{x^3})}^{10}\Rightarrow$ Coefficient of $x^5$

Generat term=

${}^{10}{C}_r(x^2{)}^{10-r}{\left(\frac{9}{x^3}\right)}^r$

Coefficient of $x^{15}$

$\Rightarrow 20-2r-3r=15$

$15=5r$

$r=1$

Coefficient of $x^{15}=$

$\Rightarrow {}^{10}{C}_1\left(a\right)$

$\Rightarrow {}^{10}{C}_r(x^2{)}^{10-r}{\left(\frac{9}{x^3}\right)}^r$

$20-2r-3r=5$

$15=5r$

$r=3$

Coefficient of $x^5=$

${}^{10}{C}_3{a}^3$

${}^{10}{C}_{1}a={}^{10}{C}_3{a}^3$

$10a=\frac{10\times 9\times 6}{6}{a}^3$

$a^2=\frac{1}{12}$

$a=\frac{1}{2\sqrt{3}}$