Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (common difference, $d$). The $n^{th}$ term is given by $T_n = a + (n-1)d$, where $a$ is the first term.
• Geometric Progression (G.P.): A sequence where the ratio between consecutive terms is constant (common ratio, $r$). The $n^{th}$ term is given by $T_n = a \cdot r^{n-1}$, where $a$ is the first term.
• Inserting Arithmetic Means: If $m$ A.Ms are inserted between $a$ and $b$, the sequence $a, A_1, \ldots, A_m, b$ forms an A.P. with $m+2$ terms. The common difference is $d = \frac{b-a}{m+1}$, and the $k^{th}$ A.M. is $A_k = a + kd$.
• Inserting Geometric Means: If $n$ G.Ms are inserted between $a$ and $b$, the sequence $a, G_1, \ldots, G_n, b$ forms a G.P. with $n+2$ terms. The common ratio is $r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$, and the $k^{th}$ G.M. is $G_k = a \cdot r^k$.

Step-by-Step Solution

Step 1: Analyze the insertion of Arithmetic Means (A.Ms) We are told that $m$ arithmetic means are inserted between $a=3$ and $b=243$. This creates an A.P. with $m+2$ terms. The common difference, $d$, of this A.P. is given by: $d = \frac{b-a}{m+1}$ Substituting the given values: $d = \frac{243 - 3}{m+1} = \frac{240}{m+1}$ The $k^{th}$ arithmetic mean is the $(k+1)^{th}$ term of the A.P. So, the $4^{th}$ A.M. ($A_4$) is the $5^{th}$ term of the A.P. (since the first term is $a$). $A_4 = a + 4d$ Substituting $a=3$ and the expression for $d$: $A_4 = 3 + 4 \left(\frac{240}{m+1}\right)$ $A_4 = 3 + \frac{960}{m+1}$

Step 2: Analyze the insertion of Geometric Means (G.Ms) We are told that three geometric means ($n=3$) are inserted between $a=3$ and $b=243$. This creates a G.P. with $3+2=5$ terms. The common ratio, $r$, of this G.P. is given by: $r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ Substituting the given values ($n=3$): $r = \left(\frac{243}{3}\right)^{\frac{1}{3+1}} = \left(81\right)^{\frac{1}{4}}$ Since $3^4 = 81$, the common ratio is: $r = 3$ The $k^{th}$ geometric mean is the $(k+1)^{th}$ term of the G.P. So, the $2^{nd}$ G.M. ($G_2$) is the $3^{rd}$ term of the G.P. $G_2 = a \cdot r^2$ Substituting $a=3$ and $r=3$: $G_2 = 3 \cdot (3)^2 = 3 \cdot 9 = 27$

Step 3: Equate the $4^{th}$ A.M. and the $2^{nd}$ G.M. and solve for $m$ The problem states that the $4^{th}$ A.M. is equal to the $2^{nd}$ G.M.: $A_4 = G_2$ Substituting the expressions derived in the previous steps: $3 + \frac{960}{m+1} = 27$ Now, we solve for $m$: Subtract 3 from both sides: $\frac{960}{m+1} = 27 - 3$ $\frac{960}{m+1} = 24$ Multiply both sides by $(m+1)$: $960 = 24(m+1)$ Divide both sides by 24: $\frac{960}{24} = m+1$ $40 = m+1$ Subtract 1 from both sides: $m = 40 - 1$ $m = 39$

Common Mistakes & Tips

• Term vs. Mean Index: Be careful not to confuse the index of a mean with the index of a term in the sequence. The $k^{th}$ A.M. is the $(k+1)^{th}$ term, and the $k^{th}$ G.M. is also the $(k+1)^{th}$ term. Use $a+kd$ and $a \cdot r^k$ for the $k^{th}$ means.
• Denominator for $d$ and exponent for $r$: When inserting $m$ means, the number of intervals between terms is $m+1$. This leads to the denominator $m+1$ for $d$ and the exponent $n+1$ for $r$ when inserting $n$ means.
• Algebraic Accuracy: Ensure that all algebraic manipulations, especially when solving for $m$, are performed correctly to avoid calculation errors.

Summary The problem involves inserting $m$ arithmetic means and 3 geometric means between 3 and 243. We first derived expressions for the $4^{th}$ arithmetic mean and the $2^{nd}$ geometric mean in terms of $m$ and the given numbers. By equating these two expressions as per the problem statement, we formed an equation that allowed us to solve for the value of $m$. The calculation yielded $m=39$.

The final answer is $\boxed{39}$.