Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. This constant is called the common difference ($d$). The $k$-th term of an AP is given by $T_k = a + (k-1)d$, where $a$ is the first term.
• Arithmetic Means: If $n$ arithmetic means are inserted between two numbers $a$ and $b$, the sequence $a, A_1, A_2, \dots, A_n, b$ forms an AP. The total number of terms in this AP is $n+2$.
• Common Difference with Means: When $n$ means are inserted between $a$ and $b$, the common difference $d$ is given by $d = \frac{b-a}{n+1}$.
• k-th Mean: The $k$-th arithmetic mean ($A_k$) is the $(k+1)$-th term of the AP, so $A_k = a + kd$.

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Step-by-Step Solution

Step 1: Setting up the Arithmetic Progression and Common Difference We are given that $n$ arithmetic means are inserted between $a$ and $100$. This creates an arithmetic progression: $a, A_1, A_2, \dots, A_n, 100$ The first term of this AP is $T_1 = a$, and the last term is $T_{n+2} = 100$. The total number of terms is $n+2$. Using the formula for the common difference $d = \frac{\text{last term} - \text{first term}}{\text{number of terms} - 1}$, we have: $d = \frac{100 - a}{(n+2) - 1} = \frac{100-a}{n+1} \quad \text{(Equation 1)}$ Why this step? This step defines the structure of the problem in terms of an AP and establishes the fundamental relationship between the first term ($a$), the last term ($100$), the number of means ($n$), and the common difference ($d$).

Step 2: Expressing the First and Last Means The first arithmetic mean, $A_1$, is the second term of the AP. Therefore, $A_1 = a + d$. The last arithmetic mean, $A_n$, is the $(n+1)$-th term of the AP. It is also the term immediately preceding the last term ($100$). Thus, $A_n = 100 - d$. Why this step? We need to express the means mentioned in the ratio condition in terms of $a$ and $d$ so we can form an algebraic equation. Using $A_n = 100 - d$ is a simplification that avoids using $A_n = a + nd$.

Step 3: Using the Ratio of the First and Last Means We are given that the ratio of the first mean to the last mean is $1:7$: $\frac{A_1}{A_n} = \frac{1}{7}$ Substitute the expressions for $A_1$ and $A_n$ from Step 2: $\frac{a+d}{100-d} = \frac{1}{7}$ Cross-multiply to eliminate denominators: $7(a+d) = 1(100-d)$ $7a + 7d = 100 - d$ Rearrange the terms to get an equation involving $a$ and $d$: $7a + 8d = 100 \quad \text{(Equation 2)}$ Why this step? This step translates the given ratio condition into a linear equation involving $a$ and $d$. This equation, along with Equation 1 and the additional condition, will form a system of equations to solve.

Step 4: Incorporating the Additional Given Condition We are given another relationship: $a + n = 33 \quad \text{(Equation 3)}$ This equation can be rewritten as $a = 33 - n$. Why this step? This provides a direct link between $a$ and $n$, allowing us to reduce the number of variables in our system of equations.

Step 5: Solving the System of Equations We have a system of three equations:

1. $d = \frac{100-a}{n+1}$
2. $7a + 8d = 100$
3. $a = 33 - n$

Substitute $a = 33 - n$ from Equation 3 into Equation 1: $d = \frac{100 - (33-n)}{n+1} = \frac{100 - 33 + n}{n+1} = \frac{67+n}{n+1} \quad \text{(Equation 4)}$ Now, substitute $a = 33 - n$ from Equation 3 into Equation 2: $7(33-n) + 8d = 100$ $231 - 7n + 8d = 100$ $8d = 100 - 231 + 7n$ $8d = 7n - 131 \quad \text{(Equation 5)}$ Now we have two equations (Equation 4 and Equation 5) with two variables, $n$ and $d$. Substitute the expression for $d$ from Equation 4 into Equation 5: $8 \left( \frac{67+n}{n+1} \right) = 7n - 131$ Multiply both sides by $(n+1)$ to clear the denominator: $8(67+n) = (7n-131)(n+1)$ Expand both sides: $536 + 8n = 7n^2 + 7n - 131n - 131$ $536 + 8n = 7n^2 - 124n - 131$ Rearrange into a standard quadratic equation ($An^2 + Bn + C = 0$): $0 = 7n^2 - 124n - 8n - 131 - 536$ $7n^2 - 132n - 667 = 0$ Solve this quadratic equation for $n$ using the quadratic formula $n = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: Here, $A=7$, $B=-132$, $C=-667$. $n = \frac{-(-132) \pm \sqrt{(-132)^2 - 4(7)(-667)}}{2(7)}$ $n = \frac{132 \pm \sqrt{17424 + 18676}}{14}$ $n = \frac{132 \pm \sqrt{36100}}{14}$ $n = \frac{132 \pm 190}{14}$ This gives two possible values for $n$: $n_1 = \frac{132 + 190}{14} = \frac{322}{14} = 23$ $n_2 = \frac{132 - 190}{14} = \frac{-58}{14}$ Since $n$ represents the number of arithmetic means, it must be a positive integer. Therefore, $n=23$ is the only valid solution. Why this step? This is the crucial step where all algebraic manipulations are performed to solve for the unknown $n$. By substituting and simplifying, we arrive at a quadratic equation, which is then solved to find the possible values of $n$. The physically meaningful solution (positive integer) is selected.

Step 6: Verification (Optional but Recommended) Let's verify if $n=23$ satisfies the given conditions. From $a+n=33$, we get $a = 33 - 23 = 10$. The common difference $d = \frac{100-a}{n+1} = \frac{100-10}{23+1} = \frac{90}{24} = \frac{15}{4}$. The first mean $A_1 = a+d = 10 + \frac{15}{4} = \frac{40+15}{4} = \frac{55}{4}$. The last mean $A_n = 100-d = 100 - \frac{15}{4} = \frac{400-15}{4} = \frac{385}{4}$. The ratio $\frac{A_1}{A_n} = \frac{55/4}{385/4} = \frac{55}{385} = \frac{1 \times 55}{7 \times 55} = \frac{1}{7}$. This matches the given ratio. The condition $a+n=33$ is $10+23=33$, which is true. Why this step? Verification ensures that the calculated value of $n$ correctly satisfies all the original conditions of the problem, increasing confidence in the answer.

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Common Mistakes & Tips

• Off-by-one errors: Carefully count the total number of terms in the AP, which is $n+2$, not $n$. This affects the calculation of the common difference.
• Algebraic errors: Be meticulous with algebraic manipulations, especially when expanding and simplifying quadratic equations. Double-check calculations.
• Interpreting the means: Remember that the first mean is the second term, and the $n$-th mean is the $(n+1)$-th term of the AP.

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Summary

The problem involves finding the number of arithmetic means ($n$) inserted between $a$ and $100$. By setting up the arithmetic progression, we derived expressions for the common difference and the means. Using the given ratio of the first and last means and the relation $a+n=33$, we formed a system of equations. Substituting and simplifying led to a quadratic equation in $n$, which we solved to find the valid positive integer value for $n$.

The final answer is $\boxed{23}$.