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

Let $m$ arithmetic mean be $A_{1}, A_{2} \ldots A_{m}$ and $G_{1}, G_{2}, G_{3}$ be

geometric mean.

The A.P. formed by arithmetic mean is,

$3, A_{1}, A_{2}, A_{3}, \ldots \ldots A_{m}, 243$

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

The G.P. formed by geometric mean

$3, G_{1}, G_{2}, G_{3}, 243$

$r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=(81)^{1 / 4}=3$

$\because A_{4}=G_{2}$

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

$\Rightarrow 3+\frac{960}{m+1}=27 \Rightarrow m+1=40 \Rightarrow m=39$