Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (common difference, $d$). For terms $T_1, T_2, T_3, T_4$, we have $T_2 - T_1 = T_3 - T_2 = T_4 - T_3 = d$. This implies $T_2 = T_1 + d$, $T_3 = T_1 + 2d$, $T_4 = T_1 + 3d$.
• Geometric Progression (G.P.): A sequence where the ratio between consecutive terms is constant (common ratio, $r$). For terms $T_1', T_2', T_3', T_4'$, we have $\frac{T_2'}{T_1'} = \frac{T_3'}{T_2'} = \frac{T_4'}{T_3'} = r$. A key property for three consecutive terms $x, y, z$ is $y^2 = xz$.
• Arithmetic Mean: The arithmetic mean of a set of numbers is their sum divided by the count of numbers. For $a, b, c$, the mean is $\frac{a+b+c}{3}$.

Step-by-Step Solution

Step 1: Express terms of the A.P. in terms of a single variable. We are given that $3, a, b, c$ are in A.P. Let the common difference be $d$. From the definition of A.P., we have: $a = 3 + d$ (1) $b = a + d = (3+d) + d = 3 + 2d$ $c = b + d = (3+2d) + d = 3 + 3d$

We can also express $b$ and $c$ in terms of $a$. From (1), $d = a - 3$. Substituting this into the expression for $b$: $b = a + (a - 3) = 2a - 3$ (2) Substituting $d$ into the expression for $c$: $c = b + d = (2a - 3) + (a - 3) = 3a - 6$ (3)

Step 2: Use the G.P. property to form an equation. We are given that $3, a-1, b+1, c+9$ are in G.P. For three consecutive terms in a G.P., the square of the middle term equals the product of the other two. Using the first three terms of the G.P.: $(a-1)^2 = 3(b+1)$

Step 3: Substitute the A.P. relationships into the G.P. equation. Substitute the expression for $b$ from equation (2) into the G.P. equation: $(a-1)^2 = 3((2a - 3) + 1)$ $(a-1)^2 = 3(2a - 2)$

Step 4: Solve the resulting quadratic equation for $a$. Expand both sides of the equation: $a^2 - 2a + 1 = 6a - 6$ Rearrange the terms to form a standard quadratic equation: $a^2 - 2a - 6a + 1 + 6 = 0$ $a^2 - 8a + 7 = 0$ Factor the quadratic equation: $(a - 1)(a - 7) = 0$ This gives two possible values for $a$: $a=1$ or $a=7$.

Step 5: Validate the possible values of $a$ by checking the G.P. terms. The terms of the G.P. are $3, a-1, b+1, c+9$.

• Case 1: If $a = 1$. The second term of the G.P. is $a-1 = 1-1 = 0$. If the second term of a G.P. is 0, and the first term is non-zero (which is 3), then the common ratio must be 0. If $r=0$, all terms after the first non-zero term must be 0. So, $a-1=0$, $b+1=0$, and $c+9=0$. Let's check if $a=1$ yields $b+1=0$ and $c+9=0$ using the A.P. relations: If $a=1$, then from equation (2), $b = 2(1) - 3 = -1$. So, $b+1 = -1+1 = 0$. This is consistent. If $a=1$, then from equation (3), $c = 3(1) - 6 = -3$. So, $c+9 = -3+9 = 6$. The G.P. terms would be $3, 0, 0, 6$. This is not a valid G.P. because if the common ratio is 0, all terms after the first zero must be zero. Here, the fourth term is 6, not 0. Therefore, $a=1$ is an extraneous solution and must be rejected.

• Case 2: If $a = 7$. The second term of the G.P. is $a-1 = 7-1 = 6$. This is non-zero, so it's a valid starting point for a G.P.

Thus, the only valid value for $a$ is $7$.

Step 6: Calculate the values of $b$ and $c$ using the valid value of $a$. Using equation (2) with $a=7$: $b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11$. Using equation (3) with $a=7$: $c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15$.

Let's verify the A.P. and G.P. with these values: A.P.: $3, 7, 11, 15$. Common difference $d=4$. This is correct. G.P.: $3, a-1, b+1, c+9 \Rightarrow 3, 7-1, 11+1, 15+9 \Rightarrow 3, 6, 12, 24$. Common ratio $r=2$. This is correct.

Step 7: Calculate the arithmetic mean of $a, b,$ and $c$. The arithmetic mean of $a, b,$ and $c$ is $\frac{a+b+c}{3}$. Substitute the calculated values: Arithmetic Mean $= \frac{7 + 11 + 15}{3}$ Arithmetic Mean $= \frac{33}{3}$ Arithmetic Mean $= 11$

Common Mistakes & Tips

• Extraneous Solutions in G.P.: Always check if any derived values for terms in a G.P. result in a zero term when it's in the denominator of the common ratio calculation or leads to an inconsistent sequence (like $3, 0, 0, 6$ for a G.P.).
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when expanding squares and rearranging terms in quadratic equations.
• Confusing A.P. and G.P. Properties: Ensure you are applying the correct definitions and formulas for each type of progression.

Summary The problem involves setting up equations based on the properties of arithmetic and geometric progressions. We first expressed $b$ and $c$ in terms of $a$ using the A.P. condition. Then, we used the G.P. condition to form an equation involving $a$ and $b$, and substituted the A.P. relationship to get a quadratic equation in $a$. Solving this quadratic yielded two potential values for $a$, one of which was rejected as it led to an inconsistent G.P. Using the valid value of $a$, we found $b$ and $c$, and finally calculated their arithmetic mean.

The final answer is \boxed{11}.