In an A.P., the sixth terms a6 = 2. If the a1 a4 a5 is the greatest, then the common difference of t

Q: Wait, why can't we just use the AM-GM inequality to maximize the product?

That is a brilliant thought! AM-GM is a powerful tool, but it only works when all the terms are guaranteed to be positive. In this problem, we do not know if $a_1$, $a_4$, and $a_5$ are positive. If any of them are negative, the product could be negative, or the AM-GM logic could fail. Calculus is the safer, more robust path here because it handles all real values of $d$.

Q: Why do we need the second derivative test? Can't we just pick the larger value of $d$?

Never assume the larger value is the maximum! In optimization, the critical points are just candidates for the peak or the valley. The second derivative test is your mathematical compass; it tells you the concavity of the curve. A negative second derivative confirms the curve is 'frowning' (concave down), which is exactly what we need for a maximum.

Q: Is there a way to solve this without expanding the whole cubic polynomial?

You could use the product rule for differentiation, but honestly, expanding it is often less prone to error than managing the product rule for three terms. Sometimes, the 'brute force' of algebra is the most reliable way to clear the path to the solution. Don't fear the expansion; embrace it as part of the journey!

Question

In an A.P., the sixth terms $a_6 = 2$. If the $a_1 a_4 a_5$ is the greatest, then the common difference of the A.P., is equal to

Options:

(A) $3/2$

(B) $8/5$

(C) $2/3$

(D) $5/8$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain, looking up at a winding path. In mathematics, especially in JEE Advanced, we often encounter problems that seem like simple sequence questions but are actually hidden optimization challenges.

We are dealing with an Arithmetic Progression (A.P.) where the sixth term is fixed at $a_6 = 2$. Our mission is to find the common difference $d$ that makes the product of the first, fourth, and fifth terms as large as possible.

### The Power of Reduction

The first step in any complex problem is to simplify the landscape. We know the general term of an A.P. is $a_n = a + (n-1)d$. Given $a_6 = 2$, we immediately write:

$$a + 5d = 2$$

By rearranging this, we find $a = 2 - 5d$. This is our anchor, allowing us to collapse a two-variable problem into a single-variable one. We are no longer juggling $a$ and $d$; we are now dancing with only $d$.

### Building the Function

Our objective is to maximize the product $P = a_1 a_4 a_5$. Let us express these terms using our new variable $d$:

$a_1 = a = 2 - 5d$

$a_4 = a + 3d = 2 - 5d + 3d = 2 - 2d$

$a_5 = a + 4d = 2 - 5d + 4d = 2 - d$

Thus, our product function becomes:

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

### The Algebraic Grind

Now, we must prepare this function for the calculus stage. While we could use the product rule, expanding the polynomial is often cleaner. Multiplying the terms:

$$f(d) = (2 - 5d)(4 - 6d + 2d^2)$$

Multiplying this out carefully, we arrive at the cubic polynomial:

$$f(d) = -10d^3 + 34d^2 - 32d + 8$$

Take a deep breath here. This is where many students rush and make a sign error. Treat each term with respect.

### The Calculus Compass

To find the maximum, we need to find where the slope of our function is zero. We differentiate $f(d)$ with respect to $d$:

$$f'(d) = -30d^2 + 68d - 32$$

Setting $f'(d) = 0$ gives us the quadratic equation $-30d^2 + 68d - 32 = 0$. Dividing by $-2$ to simplify, we get:

$$15d^2 - 34d + 16 = 0$$

Factoring this quadratic, we find $(5d - 8)(3d - 2) = 0$. This yields two critical points: $d = \frac{8}{5}$ and $d = \frac{2}{3}$.

### The Final Verification

We have two candidates for the peak. Which one is the true maximum? We use the second derivative test:

$$f''(d) = -60d + 68$$

Testing $d = \frac{8}{5}$:

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

Since $-28 < 0$, the curve is concave down, confirming a local maximum. Testing $d = \frac{2}{3}$ yields $f''(\frac{2}{3}) = 28 > 0$, which is a local minimum.

The common difference that maximizes the product is **$d = \frac{8}{5}$**.

Animated Solution for Mathematics - Sequence and Series: In an A.P., the sixth terms $a_{6} = 2$ . If the $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P., is equal to

Select Answer:

Visualized Solution

Given Information

The Product to Maximize

Single Variable Function $f (d)$

Expand for Differentiation

Differentiate $f (d)$

Set $f^{'} (d) = 0$

Critical Points

Second Derivative $f^{''} (d)$

Test $d = \frac{8}{5}$

Test $d = \frac{2}{3}$

Final Answer

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

Analyzing the Setup

The Power of Reduction

Building the Function

The Algebraic Grind

The Calculus Compass

The Final Verification

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