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:

We have to insert $$m$$ arithmetic means between $$3$$ and $$243$$. In an arithmetic progression the common difference is obtained from the formula

$$d=\dfrac{\text{last term}-\text{first term}}{\text{number of intervals}}.$$

Here the first term is $$3$$ and the last term is $$243$$. Because $$m$$ means are inserted, the total number of terms becomes $$m+2$$, so the number of equal intervals is $$m+1$$. Hence

$$d=\dfrac{243-3}{m+1}= \dfrac{240}{m+1}.$$

Now the $$4^{\text{th}}$$ arithmetic mean will lie four steps after the first term. Therefore

$$\text{4th A.M.}=3+4d=3+4\left(\dfrac{240}{m+1}\right).$$

Next we insert three geometric means between $$3$$ and $$243$$. In a geometric progression the common ratio obeys

$$\text{last term}= \text{first term}\; r^{n-1},$$

where $$n$$ is the total number of terms. With three means we have $$n=5$$, so

$$243 = 3\,r^{4}\quad\Longrightarrow\quad r^{4}=\dfrac{243}{3}=81.$$

Since $$81=3^{4}$$, we obtain

$$r = 81^{1/4}=3.$$

The geometric sequence therefore reads $$3,\,3r,\,3r^{2},\,3r^{3},\,3r^{4}$$. The $$2^{\text{nd}}$$ geometric mean is the third term overall, namely

$$\text{2nd G.M.}=3r^{2}=3(3)^{2}=3 \times 9 = 27.$$

The condition given in the problem is that the $$4^{\text{th}}$$ arithmetic mean equals the $$2^{\text{nd}}$$ geometric mean. Thus

$$3+4\left(\dfrac{240}{m+1}\right)=27.$$

Simplifying step by step, we have

$$3+\dfrac{960}{m+1}=27,$$

$$\dfrac{960}{m+1}=27-3=24,$$

$$960=24(m+1),$$

$$m+1=\dfrac{960}{24}=40,$$

$$m = 40-1 = 39.$$

So, the answer is $$39$$.