Understanding the Arithmetic Progression Problem

The problem describes four numbers, p, q, r, and s, that are in an Arithmetic Progression (AP). This means that the difference between consecutive terms is constant. Let's call this constant difference the common difference, denoted by 'd'.

The terms of the AP can be represented as:

• p
• q = p + d
• r = p + 2d
• s = p + 3d

We are given two conditions:

1. p + s = 8
2. qr = 15

Our goal is to find the difference between the largest and smallest numbers among p, q, r, and s. In an AP, the largest and smallest terms are usually the first and last terms (p and s) unless the common difference is zero (in which case all terms are equal, and the difference would be 0). Since qr = 15, neither q nor r is zero, implying the terms are not all zero. If d=0, p=q=r=s, then p+s=2p=8 gives p=4, and qr=p*p=16, which contradicts qr=15. So, d cannot be zero, and p and s are the smallest and largest terms (or vice versa).

Setting up Equations from Given Conditions

Let's substitute the AP terms in terms of p and d into the given conditions:

Condition 1: p + s = 8

Substitute s = p + 3d:

\(p + (p + 3d) = 8\)

\(2p + 3d = 8\) (Equation 1)

Condition 2: qr = 15

Substitute q = p + d and r = p + 2d:

\((p + d)(p + 2d) = 15\) (Equation 2)

Solving the System of Equations

We now have a system of two equations with two variables, p and d. We can solve this system.

From Equation 1, we can express p in terms of d:

\(2p = 8 - 3d\)

\(p = \frac{8 - 3d}{2}\)

Now substitute this expression for p into Equation 2:

\(\left(\frac{8 - 3d}{2} + d\right)\left(\frac{8 - 3d}{2} + 2d\right) = 15\)

Simplify the terms inside the parentheses:

\(\left(\frac{8 - 3d + 2d}{2}\right)\left(\frac{8 - 3d + 4d}{2}\right) = 15\)

\(\left(\frac{8 - d}{2}\right)\left(\frac{8 + d}{2}\right) = 15\)

Multiply the fractions on the left side:

\(\frac{(8 - d)(8 + d)}{4} = 15\)

Use the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\):

\(\frac{8^2 - d^2}{4} = 15\)

\(\frac{64 - d^2}{4} = 15\)

Multiply both sides by 4:

\(64 - d^2 = 15 \times 4\)

\(64 - d^2 = 60\)

Rearrange the equation to solve for \(d^2\):

\(d^2 = 64 - 60\)

\(d^2 = 4\)

Taking the square root of both sides gives the possible values for the common difference 'd':

\(d = \pm \sqrt{4}\)

\(d = \pm 2\)

Determining the Terms and the Difference

We have two possible cases for the common difference, d = 2 and d = -2.

Case 1: d = 2

Substitute d = 2 back into the equation for p:

\(p = \frac{8 - 3(2)}{2} = \frac{8 - 6}{2} = \frac{2}{2} = 1\)

Now find the other terms:

• p = 1
• q = p + d = 1 + 2 = 3
• r = p + 2d = 1 + 2(2) = 1 + 4 = 5
• s = p + 3d = 1 + 3(2) = 1 + 6 = 7

The terms are 1, 3, 5, 7. Let's check the original conditions:

• p + s = 1 + 7 = 8 (Correct)
• qr = 3 \(\times\) 5 = 15 (Correct)

In this case, the smallest number is 1 and the largest number is 7. The difference between the largest and smallest numbers is \(7 - 1 = 6\).

Case 2: d = -2

Substitute d = -2 back into the equation for p:

\(p = \frac{8 - 3(-2)}{2} = \frac{8 + 6}{2} = \frac{14}{2} = 7\)

Now find the other terms:

• p = 7
• q = p + d = 7 + (-2) = 5
• r = p + 2d = 7 + 2(-2) = 7 - 4 = 3
• s = p + 3d = 7 + 3(-2) = 7 - 6 = 1

The terms are 7, 5, 3, 1. Let's check the original conditions:

• p + s = 7 + 1 = 8 (Correct)
• qr = 5 \(\times\) 3 = 15 (Correct)

In this case, the smallest number is 1 and the largest number is 7. The difference between the largest and smallest numbers is \(7 - 1 = 6\).

In both valid cases for the common difference, the set of numbers is {1, 3, 5, 7}, just in different orders. The difference between the largest (7) and the smallest (1) number is consistently 6.

Conclusion

The difference between the largest and smallest numbers in the arithmetic progression p, q, r, s is 6.

Revision Table: Key Concepts for AP Problems

Additional Information: Solving AP Word Problems

When solving word problems involving Arithmetic Progressions, it's helpful to follow these steps:

• Identify that the sequence of numbers forms an AP (look for a constant difference).
• Define the terms of the AP using the first term (often denoted as 'a' or \(a_1\)) and the common difference 'd'. For problems with a fixed number of terms, representing them symmetrically can sometimes simplify calculations (e.g., for three terms: a-d, a, a+d; for four terms: a-3d, a-d, a+d, a+3d or simply a, a+d, a+2d, a+3d as done in this problem).
• Translate the given conditions into algebraic equations involving the first term and the common difference.
• Solve the system of equations to find the values of the first term and the common difference.
• Use these values to find the specific terms of the AP and answer the question asked in the problem.
• Always check if the calculated terms satisfy the original conditions.

This particular problem involved setting up and solving a quadratic equation for the common difference, which is a common technique in such problems.