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The sixth term of an arithmetic progression (AP) is 2. What is the value of the common difference of the AP that makes the product \(a_1a_4a_5\) the smallest?

Answer :

Final answer:

To find the value of the common difference of an AP that makes the product a₁a₄a₅ least, we can use the formula for the nth term of an AP and solve for the common difference. Taking the derivative of the product and setting it to zero will help us find the value of d that minimizes the product.

Explanation:

To find the value of the common difference of an AP that makes the product a₁a₄a₅ least, we first need to find the terms a₁, a₄, and a₅.

We are given that the sixth term of the AP is equal to 2. We can use the formula to find the nth term of an AP, which is given by aₙ = a₁ + (n-1)d, where aₙ is the nth term, a₁ is the first term, n is the number of terms, and d is the common difference.

Given that a₆ = 2, we can substitute the values into the formula: 2 = a₁ + (6-1)d. Simplifying the equation, we get 2 = a₁ + 5d.

Now, we need to find the product a₁a₄a₅.

Substituting the values into the formula, we get a₁a₄a₅ = (a₁ + 3d)(a₁ + 4d)(a₁ + 5d). We need to minimize this product, which means we need to find the value of d that will make the term d the least.

To do this, we can take the derivative of the product with respect to d and set it to zero.

Taking the derivative and setting it to zero, we get: 0 = 3a₁^2 + 30a₁d + 56d^2 + 9a₁ + 19d + 10. This is a quadratic equation in terms of d.

We can solve this equation to find the value of d that minimizes the product.

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