Given m arithmetic means (A.Ms) present between 3 and 243

$$ \therefore $$ Common difference, $$d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$$

$$ \therefore $$ 4th A.M. = a + 4d

= 3 + 4 $$ \times $$ $${{240} \over {m + 1}}$$

Also there are 3 G.M between 3 and 243

$$ \therefore $$ Common ratio (r) = $${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$$

where n = number of G.M inserted.

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

Given,

4^th A.M = 2^nd G.M

$$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$$

$$ \Rightarrow {{960} \over {m + 1}} = 24$$

$$ \Rightarrow m = 39$$