Key Concepts and Formulas

• Sum of an Arithmetic Progression (AP): The sum of the first $N$ terms of an AP with first term $a$ and common difference $d$ is given by $S_N = \frac{N}{2}[2a + (N-1)d]$.
• Algebraic Manipulation: Proficiency in simplifying equations and substituting relationships between variables is essential for solving problems involving APs.

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Step-by-Step Solution

Step 1: Express the given condition using the sum formula. We are given the condition $S_{3n} = 3S_{2n}$. We will use the formula for the sum of an AP to express $S_{3n}$ and $S_{2n}$ in terms of $a$ (the first term) and $d$ (the common difference). This will allow us to form an equation involving $a$ and $d$.

• Why this step? To translate the problem's condition into a mathematical equation that we can work with algebraically.

Using the formula $S_N = \frac{N}{2}[2a + (N-1)d]$: For $S_{3n}$: $S_{3n} = \frac{3n}{2}[2a + (3n-1)d]$ For $S_{2n}$: $S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$ Now, substitute these into the given condition $S_{3n} = 3S_{2n}$: $\frac{3n}{2}[2a + (3n-1)d] = 3 \cdot n[2a + (2n-1)d]$

Step 2: Simplify the equation to find a relationship between $a$ and $d$. We will simplify the equation from Step 1 to establish a relationship between $a$ and $d$. This relationship will be crucial for solving the problem.

• Why this step? To reduce the complexity of the equation and extract a fundamental property of the AP that links its first term and common difference, given the specific condition.

Divide both sides of the equation by $n$ (assuming $n \neq 0$, as it represents a number of terms): $\frac{3}{2}[2a + (3n-1)d] = 3[2a + (2n-1)d]$ Multiply both sides by 2 to eliminate the fraction: $3[2a + (3n-1)d] = 6[2a + (2n-1)d]$ Divide both sides by 3: $2a + (3n-1)d = 2[2a + (2n-1)d]$ Expand both sides: $2a + 3nd - d = 4a + 4nd - 2d$ Rearrange the terms to group $a$ and $d$: $2a - 4a = 4nd - 2d - 3nd + d$ $-2a = (4nd - 3nd) + (-2d + d)$ $-2a = nd - d$ $-2a = (n-1)d$ This can also be written as $2a = -(n-1)d$ or $2a = (1-n)d$. This is the key relationship between $a$ and $d$.

Step 3: Express the required ratio in terms of $a$ and $d$. We need to find the value of $\frac{S_{4n}}{S_{2n}}$. We will write the expression for $S_{4n}$ using the sum formula and then form the ratio with $S_{2n}$.

• Why this step? To set up the expression that needs to be evaluated, in preparation for substituting the relationship found in Step 2.

For $S_{4n}$ (where $N=4n$): $S_{4n} = \frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d]$ We already have $S_{2n} = n[2a + (2n-1)d]$. Now, form the ratio: $\frac{S_{4n}}{S_{2n}} = \frac{2n[2a + (4n-1)d]}{n[2a + (2n-1)d]}$ Cancel out the common factor $n$ from the numerator and denominator: $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{2a + (4n-1)d}{2a + (2n-1)d}$

Step 4: Substitute the relationship between $a$ and $d$ and calculate the value. Now, we substitute the relationship $2a = (1-n)d$ derived in Step 2 into the ratio expression from Step 3.

• Why this step? By substituting the relationship, we eliminate the variable $a$, allowing us to simplify the expression to a numerical value.

Substitute $2a = (1-n)d$ into the ratio: $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{(1-n)d + (4n-1)d}{(1-n)d + (2n-1)d}$ Factor out $d$ from the numerator and denominator: $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{d[(1-n) + (4n-1)]}{d[(1-n) + (2n-1)]}$ Assuming $d \neq 0$ (if $d=0$, the AP is constant, and the initial condition $S_{3n}=3S_{2n}$ implies $a=0$, leading to an indeterminate form $0/0$ for the ratio, but typically such problems imply non-trivial APs), we can cancel $d$: $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{1-n + 4n-1}{1-n + 2n-1}$ Combine like terms in the numerator and denominator: $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{(4n-n) + (1-1)}{(2n-n) + (1-1)}$ $\frac{S_{4n}}{S_{2n}} = 2 \cdot \frac{3n}{n}$ Cancel out the common factor $n$: $\frac{S_{4n}}{S_{2n}} = 2 \cdot 3$ $\frac{S_{4n}}{S_{2n}} = 6$

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Common Mistakes & Tips:

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when expanding brackets and combining terms. A single sign error can lead to an incorrect relationship between $a$ and $d$.
• Understanding $N$ in $S_N$: Ensure you correctly identify the number of terms ($N$) when applying the sum formula. In this problem, $N$ takes values $3n$, $2n$, and $4n$.
• Handling Special Cases: While the derivation assumes $d \neq 0$, it's good practice to consider if $d=0$ or $a=0$ leads to a valid solution. In this case, if $d=0$, the condition $S_{3n}=3S_{2n}$ implies $a=0$ (for $n \neq 0$), making the ratio $0/0$, which is usually avoided in exam problems seeking a specific numerical answer.

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Summary

The problem involves finding the ratio of sums of terms in an arithmetic progression given a relationship between other sums. The solution systematically applies the formula for the sum of an AP to express the given condition algebraically. This leads to a crucial relationship between the first term ($a$) and the common difference ($d$). This relationship is then substituted into the expression for the desired ratio, allowing for its simplification to a specific numerical value. The method highlights the importance of algebraic manipulation in solving problems related to sequences and series.

The final answer is $\boxed{6}$.