Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3 S2n, then the value

Q: Wait, why do we assume $d$ is not zero? What if the AP is constant?

That is a brilliant observation! If $d=0$, the progression is constant, and $S_n = na$. The condition $S_{3n} = 3S_{2n}$ would imply $3na = 3(2na)$, which forces $a=0$. If both $a$ and $d$ are zero, the ratio $\frac{S_{4n}}{S_{2n}}$ becomes undefined ($0/0$). In the context of JEE problems, we assume a non-trivial progression where the ratio is well-defined.

Q: Can I just pick values for $a, d,$ and $n$ to solve this quickly?

Yes, you absolutely can! This is a classic 'substitution trick' for competitive exams. If you set $n=1$, the condition becomes $S_3 = 3S_2$. You can find a relationship between $a$ and $d$ and plug them into the ratio. However, understanding the algebraic derivation is vital for when the problem is not as simple as this one.

Q: Why did we split $(4n-1)d$ into $(n-1)d + 3nd$?

We did this to force the expression $2a + (n-1)d$ to appear. Since we proved that this specific block equals zero, splitting the terms allows us to 'zero out' the complicated parts of the fraction, leaving only the terms that actually contribute to the final ratio.

Q: Is there a faster way to write the sum formula?

You could use $S_n = n \cdot a_{avg}$, where $a_{avg}$ is the average of the first and last terms. While elegant, the standard formula $S_n = \frac{n}{2}[2a + (n-1)d]$ is usually safer for algebraic manipulation in JEE problems.

Q: Does the value of $n$ affect the final answer?

Notice that in the final step, the $n$ terms cancel out completely. This tells us that the ratio is independent of $n$. Whether you take the sum of the first 3 terms or the first 300 terms, the ratio remains constant because the underlying relationship between $a$ and $d$ is fixed by the initial condition.

Question

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3n} = 3 S_{2n}$, then the value of $\frac{S_{4n}}{S_{2n}}$ is :

Options:

(A) 6

(B) 4

(C) 2

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a problem; we are peeling back the layers of an arithmetic progression to reveal the elegant symmetry hidden beneath.

Many students look at a problem like $S_{3n} = 3 S_{2n}$ and immediately reach for their pens to expand everything into a mess of variables. But wait—let us pause. In the world of competitive mathematics, the most powerful tool isn't just calculation; it is observation.

We are given the sum of the first $n$ terms of an arithmetic progression, defined by the classic formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

### The Hidden Symmetry

When we look at the condition $S_{3n} = 3 S_{2n}$, our intuition might scream, 'Expand it!' But let's look closer. Both sides contain the structure $2a + (k-1)d$.

If we substitute our formula into the given condition, we get:

$$\frac{3n}{2}[2a + (3n-1)d] = 3 \cdot \frac{2n}{2}[2a + (2n-1)d]$$

Look at the beauty of the cancellation! The factor $\frac{3n}{2}$ appears on both sides. By removing it, we simplify the entire landscape of the problem instantly:

$$2a + (3n-1)d = 2[2a + (2n-1)d]$$

### The Algebraic Alchemy

Now, we expand the right-hand side. This is where many students lose their way, but stay focused. We are hunting for a specific relationship.

Expanding gives us:

$$2a + 3nd - d = 4a + 4nd - 2d$$

By grouping the terms, we move everything to one side to see what remains:

$$4a - 2a + 4nd - 3nd - 2d + d = 0$$

$$2a + nd - d = 0$$

Factor out the $d$, and behold: $2a + (n-1)d = 0$. This is the **Key Relation**. It is the heartbeat of this problem, telling us that the sum of the first $n$ terms is actually zero.

### The Final Victory

We are asked to find the ratio $\frac{S_{4n}}{S_{2n}}$. Using our formula again, we write:

$$\frac{S_{4n}}{S_{2n}} = \frac{\frac{4n}{2}[2a + (4n-1)d]}{\frac{2n}{2}[2a + (2n-1)d]}$$

Simplifying the outer coefficients, we get $2 \cdot \frac{2a + (4n-1)d}{2a + (2n-1)d}$. Now, we use our **Key Relation** to perform a strategic split.

We rewrite $(4n-1)d$ as $(n-1)d + 3nd$ and $(2n-1)d$ as $(n-1)d + nd$. The expression becomes:

$$\frac{S_{4n}}{S_{2n}} = \frac{2[2a + (n-1)d + 3nd]}{2a + (n-1)d + nd}$$

Since $2a + (n-1)d = 0$, the terms vanish like magic, leaving us with:

$$\frac{S_{4n}}{S_{2n}} = \frac{2[0 + 3nd]}{0 + nd} = \frac{6nd}{nd} = 6$$

And there it is. The complexity collapses, the variables cancel, and we are left with the elegant integer **6**. Remember, in the JEE, the math is rarely trying to punish you; it is trying to show you a shortcut.

Animated Solution for Mathematics - Sequence and Series: Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n} = 3 S_{2 n}$ , then the value of $\frac{S _{4 n}}{S _{2 n}}$ is :

Select Answer:

Visualized Solution

Understanding the Problem

Recalling the Sum Formula

Applying the Condition

Simplifying the Equation

Expanding the Brackets

Grouping Like Terms

Finding the Key Relation

Setting up the Target Ratio

Simplifying the Target Ratio

Strategic Splitting of Terms

Final Substitution

Conclusion and Takeaway

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Hidden Symmetry

The Algebraic Alchemy

The Final Victory

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