If $a>0$ and coefficient of  $x^2,x^3,x^4$ in the expansion of ${(1 + x/a)}^6$ are in A.P., then a equals

Coefficient of $x^r$ in the expansion of $(1+x/a{)}^6$${=}^6{C}_r(1/a{)}^{\mathrm{r}}$ According to given condition  ${ }^6{C}_2(1/a{)}^2{,}^6{C}_3(1/a{)}^3{,}^6{C}_4(1/a{)}^4$ are in A.P.

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore 2\left({ }^6{C}_3\right){\left(\frac{1}{a}\right)}^3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{=}^6{C}_2{\left(\frac{1}{a}\right)}^2{+}^6{C}_4{\left(\frac{1}{a}\right)}^4\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\left(20\right)a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=15\left(a^2+1\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3a^2-8a+3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\end{array}$

$\Rightarrow a=(4+\sqrt{7})/3$ as $a>0$.