Let a1, a2, a3, ...... be an A. P. with a6 = 2. Then the common difference of this A. P., which maxi

Q: Wait, why do we reject the negative root or the other critical point? How do I know which one is the maximum?

Great question! When you set the derivative to zero, you find the 'stationary points'—the peaks and the valleys of the function. To distinguish between them, we use the second derivative test. If $P''(d) 0$, the curve is concave up, meaning you are at a valley (a minimum). Always perform this test to be certain!

Q: Can I use the AM-GM inequality here instead of calculus?

AM-GM is a powerful tool, but it requires all terms to be positive. In this problem, the terms $a_1, a_4, a_5$ depend on $d$, and they could potentially be negative depending on the value of $d$. Because we cannot guarantee positivity for all $d$, calculus is the safer, more robust, and universal method for this optimization.

Q: Is there a trick to avoid expanding the cubic polynomial?

You could use the product rule for differentiation, but for a cubic, expanding it is often less prone to error than managing the product rule with three terms. The expansion $P(d) = 2(-5d^3 + 17d^2 - 16d + 4)$ is straightforward. Sometimes, the 'long way' is actually the most reliable path to the correct answer.

Question

Let $a_1, a_2, a_3, ......$ be an A. P. with $a_6 = 2$. Then the common difference of this A. P., which maximises the product $a_1 a_4 a_5$, is :

Options:

(A) 6/5

(B) 8/5

(C) 2/3

(D) 3/2

Sigma Percentile · Accepted Answer

### The Geometry of Sequences: A Journey into Optimization

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the hidden architecture of an Arithmetic Progression (A.P.).

Imagine you are standing on a vast, rolling plain where the terrain is defined by the common difference $d$. Every step you take, every term you define, is governed by the rigid, beautiful rules of linear growth.

Our mission is to find the specific 'slope'—the common difference $d$—that allows us to reach the highest peak of the product $P = a_1 a_4 a_5$, given the constraint that the sixth term is fixed at $a_6 = 2$.

### Phase 1: The Anchor of Constraint

In any optimization problem, we must first find our anchor. We are given an A.P. where $a_6 = 2$.

Recall the fundamental definition of an A.P.: the $n$-th term is given by $a_n = a_1 + (n-1)d$. For the sixth term, this becomes:

$$a_6 = a_1 + 5d = 2$$

This is our North Star. By isolating $a_1$, we get $a_1 = 2 - 5d$.

This single substitution is the key that unlocks the entire problem. It allows us to collapse the complexity of three separate variables into a single, manageable variable: $d$.

### Phase 2: Mapping the Landscape

Now, let us express the other terms in our product, $a_4$ and $a_5$, using our new variable $d$.

The fourth term is $a_4 = a_1 + 3d$. Substituting our expression for $a_1$, we find:

$$a_4 = (2 - 5d) + 3d = 2 - 2d$$

Similarly, the fifth term is $a_5 = a_1 + 4d = (2 - 5d) + 4d = 2 - d$.

Now, look at the product $P(d) = a_1 a_4 a_5$. It is the product of three linear factors:

$$P(d) = (2 - 5d)(2 - 2d)(2 - d)$$

As we expand this, we are essentially mapping the topography of our function. After careful algebraic expansion, we arrive at the cubic function:

$$P(d) = 2(-5d^3 + 17d^2 - 16d + 4)$$

This is the 'terrain' we must navigate.

### Phase 3: The Calculus of Peaks

To find the maximum, we need to find where the slope of this terrain is zero. We invoke the power of calculus: $P'(d) = 0$.

Differentiating our cubic function with respect to $d$, we get:

$$P'(d) = 2(-15d^2 + 34d - 16) = -30d^2 + 68d - 32$$

Setting this to zero, we solve the quadratic equation $15d^2 - 34d + 16 = 0$.

Using the splitting-the-middle-term method, we factor this into $(3d - 2)(5d - 8) = 0$. This gives us two critical points:

$$d = \frac{2}{3} \quad 	ext{and} \quad d = \frac{8}{5}$$

### The Final Ascent

We have two candidates for our peak. Which one is the true maximum?

We use the second derivative test: $P''(d) = -60d + 68$. Testing our points, we find that at $d = \frac{8}{5}$:

$$P''\left(\frac{8}{5}ight) = -60\left(\frac{8}{5}ight) + 68 = -96 + 68 = -28$$

Since the second derivative is negative, the curve is concave down, confirming that **$d = \frac{8}{5}$** is indeed our local maximum.

We have successfully navigated the landscape and found the peak. Remember, in JEE Advanced, it is not just about the calculation; it is about the confidence to trust your mathematical tools. You have mastered the A.P., the algebra, and the calculus. Well done.

Animated Solution for Mathematics - Differentiation: Let $a_{1}, a_{2}, a_{3}, ......$ be an A. P. with $a_{6} = 2$ . Then the common difference of this A. P., which maximises the product $a_{1} a_{4} a_{5}$ , is :

Select Answer:

Visualized Solution

Understanding the A.P. Constraint

Relating $a_{1}$ and $d$

Expressing $a_{4}$ and $a_{5}$

Defining the Product Function $P (d)$

Applying the Maxima Condition

Differentiating $P (d)$

Solving the Quadratic Equation

Identifying the Maximum

The Sigma Insight: Maxima and Minima

The Geometry of Sequences

Phase 1

Phase 2

Phase 3

The Final Ascent

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Visualized Solution

Understanding the A.P. Constraint

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

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a4​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a5​

Defining the Product Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(d)