In an A.P., the sixth term a6 = 2. If the product a1 a4 a5 is max… — Solution | JEE Mathematics

MCQMediumJEE 2024Arithmetic Progression (AP)

In an A.P., the sixth term $a_6 = 2$ . If the product $a_1 * a_4 * a_5$ is maximized, the common difference of the A.P. is equal to:

A
$\frac{3}{2}$
B
$\frac{8}{5}$
C
$\frac{2}{3}$
D
$\frac{5}{8}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: In an A.P., $a_6 = 2$ .

Find: The common difference $d$ for which the product $a_1 a_4 a_5$ is maximized.

For an A.P.,

a_n = a_1 + (n-1)d

So,

a_6 = a_1 + 5d = 2

which gives

a_1 = 2 - 5d

Detailed Working from Extracted Solution

Now,

a_4 = a_1 + 3d = (2 - 5d) + 3d = 2 - 2d

and

a_5 = a_1 + 4d = (2 - 5d) + 4d = 2 - d

Source Conclusion and Discrepancy Note

Thus,

a_1 a_4 a_5 = (2 - 5d)(2 - 2d)(2 - d)

the solution states that after taking the derivative and setting it to zero, the maximum occurs at $d = \frac{8}{5}$ , and it concludes that the correct option is B.

Therefore, the correct option is B.

Common mistakes

Treating $a_6 = 2$ as $a_1 = 2$ . This is wrong because the sixth term in an A.P. is $a_1 + 5d$ . First write $a_1 + 5d = 2$ , then express $a_1$ in terms of $d$ .
Writing the later terms incorrectly, such as taking $a_4 = a_1 + 4d$ or $a_5 = a_1 + 5d$ . This shifts the indexing by one. Use $a_n = a_1 + (n-1)d$ carefully.
Maximizing the product without first reducing it to a single-variable function. The product must be written entirely in terms of $d$ as $f(d) = (2 - 5d)(2 - 2d)(2 - d)$ before any optimization step is attempted.

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