Question

The sum of four integers in A.P. is 24 and their product is 945 . Find the product of the smallest and greatest integers: (a) 30 (b) 27 (c) 35 (d) 39

Short answer

#Question: An arithmetic progression consists of four integers with a sum of 24 and a product of 945. Find the product of the smallest and greatest integers in the progression. a) 18 b) 27 c) 36 d) 45 #Answer: b) 27

Step-by-step solution

Set up the A.P. equation

Since we are given the sum of the four integers in A.P. as 24, let's denote the first integer as 'a' and the common difference as 'd'. Therefore, the four numbers in A.P. can be written as: a, a + d, a + 2d, a + 3d Now, the sum of these integers is equal to 24: a + (a + d) + (a + 2d) + (a + 3d) = 24

Solve for 'a' or 'd'

As we can see, this is a linear equation in 'a' and 'd', so let's simplify it and solve for one of the variables. Let's solve for 'a': 4a + 6d = 24 Divide by 4: a + \frac{3}{2}d = 6 Now, we have an equation in terms of 'a' and 'd'.

Set up the product equation

The product of the four integers in A.P. is given as 945. So, a(a + d)(a + 2d)(a + 3d) = 945

Substitute and solve for 'a'

Now, we will substitute the value of 'a' from the sum equation into the product equation: (a + \frac{3}{2}d - 6)(a + d)(a + 2d)(a + 3d) = 945 Now, since we have a single equation in terms of d, we can solve for d by trial and error method. We try a few values of d: d = 2: a = 3 the A.P. is 3, 5, 7, 9 Now, let's check the product of these integers: 3 × 5 × 7 × 9 = 945 Thus, 'a' is 3, and 'd' is 2. So, the smallest integer is 3, and the largest integer is 9.

Find the product of the smallest and largest integers

Now that we have the smallest and largest integers, we can find their product: 3 × 9 = 27 So, the product of the smallest and greatest integers is 27. Hence, the correct answer is (b) 27.

Key concepts

Sum of terms in arithmetic progression

Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference, denoted as \(d\). In our problem, we have four integers in A.P. whose sum is given as 24. Let’s denote the first term as \(a\). Thus, the integers can be expressed as:

• \(a\)
• \(a + d\)
• \(a + 2d\)
• \(a + 3d\)

To find the sum of these terms, we combine them: \(a + (a + d) + (a + 2d) + (a + 3d) = 24\). This simplifies to: \(4a + 6d = 24\).
The equation \(a + \frac{3}{2}d = 6\) results after dividing by 4. It indicates that the sum can be split into smaller linear components of \(a\) and \(d\). Understanding this equation is crucial for solving such problems as it allows you to express one variable in terms of the other. This step is the foundation for further calculations, such as finding the product of these integers.

Product of terms in arithmetic progression

The task asks us to analyze the product of four integers in Arithmetic Progression which equates to 945. With \(a\) as the first term and \(d\) as the common difference, the product can be represented as: \(a(a + d)(a + 2d)(a + 3d) = 945\). This equation emphasizes how the multiplication of terms in an A.P. leads us to a specific integer value, in this case, 945. By substituting \(a\) from the sum equation into this product equation, we transform it to be solely dependent on \(d\). Exploring different trial values for \(d\), starting with \(d = 2\) results in \(a = 3\), producing the sequence: 3, 5, 7, 9. Confirm the product: 3 × 5 × 7 × 9 = 945, validating our estimation of \(d\) and \(a\). The smallest integer is 3 and the largest is 9, establishing a solution pathway towards understanding broader principles of A.P. contributions to total product calculations.

Linear equations in two variables

The interplay of linear equations in two variables is often encountered in Arithmetic Progressions when working with sums or products of sequences. In our example, the sum equation \(4a + 6d = 24\) transforms into \(a + \frac{3}{2}d = 6\). This is a linear equation involving two variables: \(a\) and \(d\). Such equations are fundamental to finding solutions efficiently. They allow us to handle multiple unknowns in structured mathematical processes. To solve for these variables, breaking the equation down isolates one variable for substitution into another equation—streamlining the solving process. For example, by solving for \(a\) in terms of \(d\), as in \(a = 6 - \frac{3}{2}d\), and then substituting it in other equations, it becomes much easier to navigate through problems involving terms in arithmetic progression. These interactions are pivotal in mathematical reasoning and problem-solving, ensuring all parts of the equation map correctly back to each concept.