QJEEM20Ma7078: JEE Main 2020 Mathematics - Class 11 JEEM20Ma7078 Solution: If m arithmetic means (A.Ms) and three geometric means (G.Ms)…

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

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

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

$\Rightarrow m = 39$