If m arithmetic means (A.Ms) and three geometric means (G.Ms)

$3, \mathrm{~A}_{1}, \mathrm{~A}_{2} \ldots \ldots \ldots \mathrm{A}_{\mathrm{m}}, 243$

$\mathrm{d}=\frac{243-3}{\mathrm{~m}+1}=\frac{240}{\mathrm{~m}+1}$

Now $3, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, 243$

$r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3$

$\therefore \quad \mathrm{A}_{4}=\mathrm{G}_{2}$

$\Rightarrow \quad a+4 d=a r^{2}$

$3+4\left(\frac{240}{m+1}\right)=3(3)^{2}$

$\mathrm{m}=39$