If a₁, a₂, a₃, ..., aₙ are in A.P. and a₁ + a₄ + a₇ + ... + a₁₆ = 114, then a₁ + a₆ + a₁₁ + a₁₆ is e

Q: Wait, why can't I just find the first term $a$ and the common difference $d$?

You absolutely could! You would set up a system of equations, solve for $a$ and $d$, and then calculate the target sum. However, in competitive exams like JEE, time is your most precious resource. Recognizing the symmetry property allows you to bypass the algebra entirely, saving you minutes of calculation and reducing the chance of a silly arithmetic error.

Q: How do I know for sure that $a_6 + a_{11}$ is equal to $a_1 + a_{16}$?

Great question! It all comes down to the indices. In an A.P., the term $a_n$ is defined as $a + (n-1)d$. If you add two terms $a_m$ and $a_n$, you get $2a + (m+n-2)d$. Notice that the sum depends only on the sum of the indices $(m+n)$. Since $6+11 = 17$ and $1+16 = 17$, the sum of the terms must be identical.

Q: Is this property applicable to every Arithmetic Progression?

Yes, it is a universal property of Arithmetic Progressions! Whether the common difference is positive, negative, or even zero, the symmetry of the sequence ensures that equidistant terms from the ends always sum to the same value. It is one of the most elegant tools in your mathematical toolkit.

Question

If $a_1, a_2, a_3, \dots, a_n$ are in A.P. and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to :

Options:

(A) 38

(B) 98

(C) 76

(D) 64

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a long, straight road, and every few meters, there is a milestone. These milestones represent an Arithmetic Progression (A.P.), a sequence where the gap between any two consecutive points is always the same.

In our problem, we are given a sequence $a_1, a_2, a_3, \dots, a_n$ and a specific sum:
$$a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114$$

At first glance, this looks like a standard algebra problem where you might be tempted to dive into the formulas for the $n$-th term, $a_n = a + (n-1)d$. However, before you start scribbling down equations for $a$ and $d$, let us pause and look for the hidden geometry of the sequence.

### The Power of Pairs

There is a beautiful, almost poetic property in an A.P.: the sum of terms equidistant from the beginning and the end is constant. Mathematically, this means $a_k + a_{n-k+1} = a_1 + a_n$.

The magic lies in the indices. If the sum of the indices is constant, the sum of the terms is constant. Let us look at our given terms: $a_1, a_4, a_7, a_{10}, a_{13}, a_{16}$.

If we pair them up, we get:
*   Pair 1: $(a_1 + a_{16})$, where $1+16=17$.
*   Pair 2: $(a_4 + a_{13})$, where $4+13=17$.
*   Pair 3: $(a_7 + a_{10})$, where $7+10=17$.

Because the sum of the indices for every pair is exactly $17$, the property tells us that $a_1 + a_{16} = a_4 + a_{13} = a_7 + a_{10}$.

### Simplifying the Equation

Now, the problem transforms from a tedious calculation into a simple substitution. We can rewrite our original sum as:
$$(a_1 + a_{16}) + (a_4 + a_{13}) + (a_7 + a_{10}) = 114$$

Since all these pairs are equal to $(a_1 + a_{16})$, we can replace them all:
$$(a_1 + a_{16}) + (a_1 + a_{16}) + (a_1 + a_{16}) = 114$$

This simplifies beautifully to:
$$3(a_1 + a_{16}) = 114$$

Dividing by $3$, we find that $a_1 + a_{16} = 38$. This value is the key to the entire puzzle.

### The Final Leap

We are asked to find the sum $S = a_1 + a_6 + a_{11} + a_{16}$. Let us apply our pairing strategy again.

We group them as $(a_1 + a_{16}) + (a_6 + a_{11})$. We already know $a_1 + a_{16} = 38$.

What about $a_6 + a_{11}$? The sum of the indices is $6 + 11 = 17$. By our symmetry property, $a_6 + a_{11}$ must also equal $a_1 + a_{16}$, which is $38$.

Therefore, our target sum is simply:
$$S = 38 + 38 = 76$$

By observing the symmetry rather than brute-forcing the algebra, we have arrived at the answer with elegance and speed. This is the heart of JEE Advanced mathematics: finding the simplest path through the complexity. The final answer is **76**.

Animated Solution for Mathematics - Sequence and Series: If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in A.P. and $a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to :

Select Answer:

Visualized Solution

Introduction to the A.P. Sequence

Property of Equidistant Terms

Visualizing the Pairs

Grouping the Given Equation

Equating the Pairs

Substituting the Pairs

Solving for the Constant Sum

Analyzing the Target Sum

Applying the Property to the Target

Final Substitution

The Final Answer

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

Analyzing the Setup

The Power of Pairs

Simplifying the Equation

The Final Leap

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