Let S₁ be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n term

Q: Wait, why can't we just solve for $a$, $d$, and $n$?

Great question! In many JEE problems, you are given fewer equations than variables. This is intentional. It forces you to look for a 'block' or 'structure' that cancels out, rather than brute-forcing individual values. The problem is mathematically designed to be solved as a whole.

Q: Why did the $n$ terms simplify so cleanly?

It's not magic; it's the linearity of the sum formula. Because the sum of an AP is a quadratic in $n$, the difference between two sums of an AP is also a structured expression. Trust the algebra! The coefficients of $a$ and $d$ are designed to align perfectly.

Q: Is there a faster way to do this?

You could use the property that the sum of an AP is $S_n = An^2 + Bn$. But honestly, the algebraic method we used is the most robust and least prone to errors. Stick to the fundamentals, and the structure will reveal itself!

Question

Let $S_1$ be the sum of first $2n$ terms of an arithmetic progression. Let $S_2$ be the sum of first $4n$ terms of the same arithmetic progression. If $(S_2 - S_1)$ is $1000$, then the sum of the first $6n$ terms of the arithmetic progression is equal to:

Options:

(A) $1000$

(B) $7000$

(C) $5000$

(D) $3000$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, multi-layered puzzle. You have an Arithmetic Progression (AP), but you do not know its starting point $a$, its common difference $d$, or even the number of terms $n$.

Yet, you are asked to find the sum of the first $6n$ terms. It feels impossible, but the secret of JEE Advanced is to look for the structure of the whole when the individual pieces remain elusive.

### The Toolkit: The Sum Formula

Our journey begins with the fundamental tool for the sum of the first $k$ terms of an AP:

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

We are given $S_1$ (the sum of $2n$ terms) and $S_2$ (the sum of $4n$ terms). We express these as:

$$S_1 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$$

$$S_2 = \frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d]$$

### The Algebraic Dance

We face the condition $S_2 - S_1 = 1000$. Substituting our expressions, we get:

$$2n[2a + (4n-1)d] - n[2a + (2n-1)d] = 1000$$

Distributing the terms, we obtain:

$$4na + 2n(4n-1)d - 2na - n(2n-1)d = 1000$$

Grouping the $a$ terms yields $2na$. For the $d$ terms, we factor out $nd$:

$$nd[2(4n-1) - (2n-1)] = 1000$$

Simplifying the bracketed expression:

$$nd[8n - 2 - 2n + 1] = nd[6n - 1]$$

Our equation now stands as $2na + nd(6n-1) = 1000$. Factoring out $n$, we reach the revelation:

$$n[2a + (6n-1)d] = 1000$$

### The Final Act

We are asked to find $S_{6n}$. Writing its formula:

$$S_{6n} = \frac{6n}{2}[2a + (6n-1)d] = 3n[2a + (6n-1)d]$$

Look closely at this expression; it is exactly $3$ times the expression we just derived. Since $n[2a + (6n-1)d] = 1000$, it follows that:

$$S_{6n} = 3 \cdot 1000 = 3000$$

We did not need to find $a$, $d$, or $n$. By recognizing the pattern, we arrive at the **final answer of 3000**.

Select Answer:

Visualized Solution

Defining the Given Information

The Sum Formula for an A.P.

Expressions for $S_{1}$ and $S_{2}$

Setting up $S_{2} - S_{1} = 1000$

Expanding the Terms

Grouping Like Terms

Simplifying the $d$ Coefficient

Extracting the Key Relation

Defining the Target Sum $S_{6 n}$

Connecting the Pieces

Final Calculation

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

Analyzing the Setup

The Toolkit

The Algebraic Dance

The Final Act

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