Let a₁ = 8, a₂, a₃, ..., a_n be an A.P. If the sum of its first four terms is 50 and the sum of its

Q: Wait, why do we use $a_{n-3}$ for the last four terms? Shouldn't it be $a_{n-4}$?

That is a great question! Let's count carefully. If the sequence has $n$ terms, the last term is $a_n$. The term before it is $a_{n-1}$, then $a_{n-2}$, and the fourth one from the end is $a_{n-3}$. If you used $a_{n-4}$, you would be looking at the fifth term from the end. Always visualize the sequence: $a_n, a_{n-1}, a_{n-2}, a_{n-3}$ are the last four.

Q: How do I know if there is one middle term or two?

The number of middle terms depends on whether $n$ is odd or even. If $n$ is odd, there is exactly one middle term at position $\frac{n+1}{2}$. If $n$ is even, as in our case where $n=14$, there are two middle terms at positions $\frac{n}{2}$ and $\frac{n}{2} + 1$. Since $14$ is even, we look for the $7^{th}$ and $8^{th}$ terms.

Q: Can I use the sum formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ to solve this?

You could, but it would be overkill! The sum formula is perfect for the sum of *all* $n$ terms. Here, we are only given the sum of specific subsets (the first four and the last four). Using the general term formula $a_k = a_1 + (k-1)d$ is much more direct and less prone to algebraic errors for small subsets of terms.

Question

Let $a_1 = 8, a_2, a_3, \dots, a_n$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Sigma Percentile · Accepted Answer

### The Elegance of Arithmetic Progressions

Imagine you are standing at the base of a grand, infinite staircase. Each step you take represents a term in an Arithmetic Progression (A.P.).

You know exactly where you started—the first step, $a_1$, is at a height of $8$. You also know that every time you move to the next step, you climb by a constant, fixed height, which we call the common difference, $d$.

This is the heartbeat of an A.P.: a predictable, rhythmic climb. Today, we are going to decode the structure of this staircase using only the clues provided by the sum of its parts.

### Phase 1: Decoding the Start

We are told that the sum of the first four steps is $50$. Let's translate this into the language of algebra.

The first four terms are $a_1, a_2, a_3,$ and $a_4$. In an A.P., we can express any term using the first term and the common difference: $a_k = a_1 + (k-1)d$.

Thus, our sum becomes:

$$a_1 + (a_1+d) + (a_1+2d) + (a_1+3d) = 50$$

Simplifying this, we get $4a_1 + 6d = 50$. Since we know $a_1 = 8$, we substitute it in: $4(8) + 6d = 50$, which simplifies to $32 + 6d = 50$.

Subtracting $32$ from both sides gives us $6d = 18$, and finally, $d = 3$. We have successfully unlocked the rhythm of our staircase; every step is exactly $3$ units higher than the last.

### Phase 2: The Bridge to the End

Now, let's look at the other end of the sequence. The problem tells us that the sum of the last four terms is $170$.

These terms are $a_{n-3}, a_{n-2}, a_{n-1},$ and $a_n$. Using our general term formula, we express each of these in terms of $n$ and $d$:

$$a_{n-3} = a_1 + (n-4)d$$
$$a_{n-2} = a_1 + (n-3)d$$
$$a_{n-1} = a_1 + (n-2)d$$
$$a_n = a_1 + (n-1)d$$

When we sum these, we get $4a_1 + (4n-10)d = 170$. Substituting $a_1 = 8$ and $d = 3$, we have $4(8) + (4n-10)3 = 170$.

Expanding this, we get $32 + 12n - 30 = 170$, which simplifies beautifully to $12n + 2 = 170$. Solving for $n$, we find $12n = 168$, so $n = 14$. Our staircase has exactly $14$ steps.

### Phase 3: Finding the Middle

With $n = 14$, we are dealing with an even number of terms. In such a sequence, there isn't just one middle term; there are two.

They sit at positions $\frac{n}{2}$ and $\frac{n}{2} + 1$. For $n = 14$, these are the $7^{th}$ and $8^{th}$ terms, $a_7$ and $a_8$.

Let's calculate them:

$$a_7 = a_1 + 6d = 8 + 6(3) = 8 + 18 = 26$$
$$a_8 = a_1 + 7d = 8 + 7(3) = 8 + 21 = 29$$

### The Final Celebration

We have reached the summit! The question asks for the product of these two middle terms.

We simply multiply $26$ and $29$:

$$	ext{Product} = 26 	imes 29 = 754$$

And there it is. By breaking the problem into manageable phases—first finding the common difference, then the total number of terms, and finally identifying the middle—we have navigated the complexity with ease.

Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the hidden geometric or algebraic structure and letting it guide you to the solution. The final answer is **754**.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Enter Numerical Value:

Visualized Solution

Visualizing the A.P. Sequence

Sum of First Four Terms

Substituting $a_{1}$ to find $d$

Solving for Common Difference $d$

Sum of Last Four Terms

Expanding the Last Terms

Setting up the Equation for $n$

Solving for Number of Terms $n$

Identifying Middle Terms

Calculating the Seventh Term $a_{7}$

Calculating the Eighth Term $a_{8}$

Final Product Calculation

Summary and Key Takeaway

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

The Elegance of Arithmetic Progressions

Phase 1

Phase 2

Phase 3

The Final Celebration

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Enter Numerical Value:

Visualized Solution

Visualizing the A.P. Sequence

Sum of First Four Terms

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solving for Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Sum of Last Four Terms

Expanding the Last Terms

Setting up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Solving for Number of Terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n