If a distance of 50m is covered in 1 minute, that 90m in 2 minutes and 130m in 3 minutes find the distance covered in 15 minutes.

Options

A. 610m
B. 750m
C. 1000m
D. 650m

Answer

To solve the problem, we need to determine the distance covered in 15 minutes based on the distances covered in the first three minutes. We can observe that the distances form an arithmetic progression (AP). Let's break down the solution step by step.

### Step 1: Identify the distances covered in the first three minutes.
- In 1 minute: 50 meters
- In 2 minutes: 90 meters
- In 3 minutes: 130 meters

### Step 2: Calculate the common difference (D) of the AP.
To find the common difference, we can subtract the distance covered in the previous minute from the distance covered in the current minute:
- From 1 minute to 2 minutes: \(90 - 50 = 40\) meters
- From 2 minutes to 3 minutes: \(130 - 90 = 40\) meters

Thus, the common difference \(D = 40\) meters.

### Step 3: Identify the first term (A) of the AP.
The first term \(A\) is the distance covered in the first minute, which is:
- \(A = 50\) meters

### Step 4: Use the formula for the nth term of an AP to find the distance covered in 15 minutes.
The formula for the nth term of an arithmetic progression is given by:
\[ A_n = A + (n - 1) \times D \]
Where:
- \(A_n\) is the nth term (distance covered in n minutes),
- \(A\) is the first term,
- \(n\) is the term number,
- \(D\) is the common difference.

For our case:
- \(n = 15\)
- \(A = 50\)
- \(D = 40\)

Substituting these values into the formula:
\[ A_{15} = 50 + (15 - 1) \times 40 \]
\[ A_{15} = 50 + 14 \times 40 \]
\[ A_{15} = 50 + 560 \]
\[ A_{15} = 610 \text{ meters} \]

### Final Answer:
The distance covered in 15 minutes is **610 meters**.

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