If the sum of the first 2n terms of the A.P. 2, 5, 8, ... is equal to the sum of the first n terms o

Q: Wait, why do we use $2n$ for the first sequence and $n$ for the second? Is that a typo?

It is not a typo! In JEE problems, the constraints are everything. The problem explicitly states 'sum of the first $2n$ terms' for the first A.P. and 'sum of the first $n$ terms' for the second. Always read the problem statement twice before starting your calculation. This is a deliberate constraint to test your attention to detail.

Q: Can I just cancel $n$ from both sides of the equation? Is that mathematically sound?

Yes, absolutely. Since $n$ represents the number of terms in a sequence, $n$ must be a positive integer ($n \in \mathbb{Z}^+$). Because $n > 0$, it is definitely not zero, so dividing both sides by $n$ is a perfectly valid algebraic operation that simplifies your work immensely.

Q: Why can't we just use energy conservation or some other physics shortcut?

This is a pure mathematics problem involving Arithmetic Progressions. While physics intuition is great, here we must rely on the algebraic properties of sequences. The 'shortcut' is simply being systematic with the sum formula and simplifying the equation before expanding the brackets.

Question

If the sum of the first $2n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$, then $n$ equals

Options:

(A) 10

(B) 12

(C) 11

(D) 13

Sigma Percentile · Accepted Answer

### The Balancing Act: A Journey Through Arithmetic Progressions

Welcome, future engineer! Today, we are going to tackle a problem that might seem like a simple algebraic exercise, but it is actually a beautiful lesson in balance and systematic thinking. We are dealing with two distinct Arithmetic Progressions (A.P.s) and a condition that forces them to meet at a specific point.

### Phase 1: Defining the Players

Imagine you are standing in front of a giant balance scale. On the left pan, we have the sum of the first $2n$ terms of the sequence $2, 5, 8, \dots$. On the right pan, we have the sum of the first $n$ terms of the sequence $57, 59, 61, \dots$.

The problem states these two pans are perfectly level. Our goal is to find the value of $n$ that makes this equilibrium possible.

First, let us extract the DNA of these sequences. For the first A.P., the first term $a_1 = 2$ and the common difference $d_1 = 5 - 2 = 3$. We are summing $2n$ terms here.

For the second A.P., the first term $a_2 = 57$ and the common difference $d_2 = 59 - 57 = 2$. We are summing $n$ terms here. Note the difference in the number of terms—this is the pivot point of our problem.

### Phase 2: The Power of the Sum Formula

To solve this, we need our most reliable tool: the sum formula for an A.P., $S_k = \frac{k}{2}[2a + (k-1)d]$. This formula is the bridge between the individual terms and the total accumulation.

We will apply this to both sides of our balance scale. For the left side, with $2n$ terms, we have:

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

For the right side, with $n$ terms, we have:

$$S_n = \frac{n}{2}[2(57) + (n-1)2]$$

### Phase 3: The Algebraic Dance

Now, we equate them. This is where many students get lost in the algebra. Stay calm and look for patterns. We have:

$$\frac{2n}{2}[2(2) + (2n-1)3] = \frac{n}{2}[2(57) + (n-1)2]$$

Let us simplify the left side first. The $2$ in the fraction $\frac{2n}{2}$ cancels out, leaving us with $n$. Inside the bracket, we have $4 + 6n - 3$, which simplifies to $6n + 1$. So, the left side is $n(6n + 1)$.

Now, the right side. We keep the $\frac{n}{2}$ for a moment. Inside the bracket, we have $114 + 2n - 2$, which simplifies to $112 + 2n$. So, the right side is $\frac{n}{2}(112 + 2n)$.

### Phase 4: The Elegant Cancellation

We are left with the equation:

$$n(6n + 1) = \frac{n}{2}(112 + 2n)$$

Since $n$ is the number of terms, we know $n 
eq 0$. This allows us to divide both sides by $n$, effectively removing it from the equation. This is the moment of clarity! We are left with:

$$6n + 1 = \frac{1}{2}(112 + 2n)$$

Distributing the $\frac{1}{2}$ on the right side gives us $56 + n$. Now, the equation is a simple linear one:

$$6n + 1 = 56 + n$$

Subtracting $n$ from both sides and $1$ from both sides, we get $5n = 55$. Finally, dividing by $5$, we find **$n = 11$**.

### Conclusion

There you have it! By breaking the problem down into its fundamental components and systematically simplifying, we turned a potentially intimidating equation into a straightforward solution.

Remember, in JEE Advanced, the complexity often lies in the setup, not the final calculation. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math.

Animated Solution for Mathematics - Sequence and Series: If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Select Answer:

Visualized Solution

The Balancing Act of Two A.P.s

Parameters of the First A.P.

Parameters of the Second A.P.

The Sum Formula

Setting up the Equation

Simplifying the Left Side

Simplifying the Right Side

Equating and Cancelling $n$

Eliminating the Fraction

Solving for $n$

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

The Balancing Act

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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