If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4^th A.M. is equal to 2^nd G.M., then m is equal to _________ .

Given m arithmetic means (A.Ms) present between 3 and 243<br><br>$\therefore$ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$<br><br>$\therefore$ 4th A.M. = a + 4d<br><br>= 3 + 4 $\times$ ${{240} \over {m + 1}}$<br><br>Also there are 3 G.M between 3 and 243<br><br>$\therefore$ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$<br><br>where n = number of G.M inserted.<br><br>$\therefore$ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$<br><br>Given, <br><br>4^th A.M = 2^nd G.M<br><br>$\Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$<br><br>$\Rightarrow {{960} \over {m + 1}} = 24$<br><br>$\Rightarrow m = 39$