Let a_1, a_2, ..., a_2024 be an Arithmetic Progression such that a_1 + (a_5 + a_10 + a_15 + ... + a_

Q: Wait, why is the sum of equidistant terms always constant in an A.P.?

That is a fantastic question! Think of an A.P. as a sequence where each term increases by a fixed common difference $d$. If you take the first term $a_{1}$ and the last term $a_{n}$, their sum is $a_{1} + (a_{1} + (n-1)d) = 2a_{1} + (n-1)d$. Now, take the second term $a_{2}$ and the second-to-last term $a_{n-1}$. That is $(a_{1} + d) + (a_{1} + (n-2)d) = 2a_{1} + (n-1)d$. They are identical! This symmetry is the backbone of many A.P. problems.

Q: How do I know for sure that the terms in the bracket pair up perfectly?

Great observation. We have terms $a_{5}, a_{10}, \dots, a_{2020}$. These are $404$ terms. When you pair the first ($a_{5}$) and the last ($a_{2020}$), the sum of their indices is $5 + 2020 = 2025$. The next pair is $a_{10} + a_{2015}$, and $10 + 2015 = 2025$. Since the sum of indices is always $2025$, and the total number of terms is even ($404$), every single term finds a partner. It is a perfect, balanced structure.

Q: Can I solve this by finding $a_{1}$ and $d$ individually?

Technically, yes, but it is like trying to cut a piece of paper with a chainsaw! You would have two unknowns ($a_{1}$ and $d$) and only one equation. You would be stuck. The beauty of this problem is that you do not need to know the individual values of $a_{1}$ or $d$. You only need the value of the sum $a_{1} + a_{2024}$. Always look for the 'shortcut' that uses the properties of the sequence rather than brute-forcing the variables.

Question

Let $a_{1},a_{2},...,a_{2024}$ be an Arithmetic Progression such that $a_{1}+(a_{5}+a_{10}+a_{15}+...+a_{2020})+a_{2024}=2233$. Then $a_{1}+a_{2}+a_{3}+...+a_{2024}$ is equal to

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Arithmetic Progressions

Imagine you are standing before a massive, daunting equation: 
$$a_{1} + (a_{5} + a_{10} + \dots + a_{2020}) + a_{2024} = 2233$$

It looks like a mountain of variables, but in the world of JEE Advanced, the most intimidating problems often hide the most elegant, simple truths. Today, we are going to peel back the layers of this arithmetic progression and discover the secret pattern that makes this problem collapse into a simple, beautiful solution.

### The Secret Weapon: Equidistant Terms

Before we touch the algebra, let's talk about the soul of an Arithmetic Progression (A.P.). There is a fundamental property that every top-tier student must keep in their toolkit: the sum of terms equidistant from the beginning and the end is always constant.

Mathematically, for an A.P. with $n$ terms:
$$a_{k} + a_{n-k+1} = a_{1} + a_{n}$$

Why does this happen? As you move forward from the start, you add the common difference $d$, and as you move backward from the end, you subtract that same $d$. They cancel out perfectly! In our case, with $n = 2024$, this means $a_{1} + a_{2024} = a_{k} + a_{2025-k}$. This is our key to the kingdom.

### Decoding the Bracket

Now, let's look at that intimidating bracket: $(a_{5} + a_{10} + \dots + a_{2020})$. The indices are $5, 10, 15, \dots, 2020$. These are all multiples of $5$.

To find out how many terms are hiding in there, we simply divide the last index by $5$:
$$\frac{2020}{5} = 404$$

So, we have exactly $404$ terms. Now, let's pair them up. We pair the first term, $a_{5}$, with the last, $a_{2020}$. The sum of their indices is $5 + 2020 = 2025$.

We pair the second, $a_{10}$, with the second-to-last, $a_{2015}$. The sum of their indices is $10 + 2015 = 2025$. Do you see the magic? Every single pair sums to the same value as $a_{1} + a_{2024}$. Since we have $404$ terms, we have exactly:
$$\frac{404}{2} = 202 	ext{ such pairs.}$$

### The Synthesis

Let's define our constant sum as $S = a_{1} + a_{2024}$. Our original equation can now be rewritten:
$$a_{1} + (a_{5} + a_{10} + \dots + a_{2020}) + a_{2024} = 2233$$

The outer terms $a_{1} + a_{2024}$ give us one $S$. The bracketed part, which consists of $202$ pairs, each equal to $S$, gives us $202S$.

So, the equation simplifies to:
$$S + 202S = 2233 \Rightarrow 203S = 2233$$

Dividing $2233$ by $203$ gives us $S = 11$. Just like that, the mountain has been leveled.

### The Grand Finale

We are almost there. The question asks for the sum of all $2024$ terms, denoted as $S_{2024}$. The formula for the sum of an A.P. is:
$$S_{n} = \frac{n}{2}(a_{1} + a_{n})$$

Substituting our values, we get:
$$S_{2024} = \frac{2024}{2}(a_{1} + a_{2024})$$

We know $a_{1} + a_{2024} = S = 11$. Thus:
$$S_{2024} = 1012 	imes 11 = 11132$$

Performing this final multiplication, we arrive at **$11132$**. By trusting the symmetry and looking for the pattern, we turned a terrifying problem into a moment of pure mathematical joy. Keep this perspective, and you will conquer any problem the exam throws at you!

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, ..., a_{2024}$ be an Arithmetic Progression such that $a_{1} + (a_{5} + a_{10} + a_{15} + ... + a_{2020}) + a_{2024} = 2233$ . Then $a_{1} + a_{2} + a_{3} + ... + a_{2024}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

The Property of Equidistant Terms

Identify Terms in the Bracket

Count the Number of Terms

Pairing the Equidistant Terms

Number of Pairs

Define the Constant Sum $S$

Simplify and Solve for $S$

The Sum of the Entire A.P.

Final Calculation

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

The Hidden Symmetry of Arithmetic Progressions

The Secret Weapon

Decoding the Bracket

The Synthesis

The Grand Finale

Similar Questions

Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57,$ then $S_{30} - S_{10}$ is equal to:

The sum of the common terms of the following three arithmetic progressions. $3, 7, 11, 15, \dots, 399$ , $2, 5, 8, 11, \dots, 359$ and $2, 7, 12, 17, \dots, 197$ , is equal to

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first $4 n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$ , then the sum of the first $6 n$ terms of the arithmetic progression is equal to:

Let $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms. Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$ . If $A_{3} = - 153, A_{5} = - 435$ and $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66$ , then $a_{17} - A_{7}$ is equal to

Let $A P (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $A P (1; 3) \cap A P (2; 5) \cap A P (3; 7) = A P (a; d)$ then $a + d$ equals

If $lo g_{3} 2, lo g_{3} (2^{x} - 5), lo g_{3} (2^{x} - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to

Let $3, 7, 11, 15, \dots, 403$ and $2, 5, 8, 11, \dots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to