Question:
If $m$ arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that $4^{\text {th }}$ A.M. is equal to $2^{\text {nd }}$ G.M., then $\mathrm{m}$ is equal to______________
Solution:
$3, \mathrm{~A}_{1}, \mathrm{~A}_{2} \ldots \ldots \ldots \mathrm{A}_{\mathrm{m}}, 243$
$\mathrm{d}=\frac{243-3}{\mathrm{~m}+1}=\frac{240}{\mathrm{~m}+1}$
Now $3, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, 243$
$r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3$
$\therefore \quad \mathrm{A}_{4}=\mathrm{G}_{2}$
$\Rightarrow \quad a+4 d=a r^{2}$
$3+4\left(\frac{240}{m+1}\right)=3(3)^{2}$
$\mathrm{m}=39$