Question

Let A1, A2, A3, ..., A49 be 49 arithmetic means between 49 and 149. Then the mean of A1, A25, A47 and A49 is:

• A. 110
• B. 120
• C. 130
• D. 140

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When \( n \) Arithmetic Means (AMs) are inserted between two numbers \( a \) and \( b \), all the numbers (including \( a \) and \( b \)) form an Arithmetic Progression. We can use the properties of symmetry in an A.P. to find the mean of specific terms.
Step 2: Key Formula or Approach:
1. Total terms in A.P. = \( 49 + 2 = 51 \).
2. Let \( a = 49 \) and \( b = 149 \). The common difference \( d = \frac{b-a}{n+1} \).
3. In an A.P., \( a_k + a_{n-k+1} = a_1 + a_n \).
Step 3: Detailed Explanation:
1. The sequence is \( 49, A_1, A_2, \dots, A_{49}, 149 \). 2. The common difference \( d = \frac{149 - 49}{49 + 1} = \frac{100}{50} = 2 \). 3. The terms we need are: \[ A_1 = 49 + 2 = 51 \] \[ A_{25} = 49 + 25(2) = 49 + 50 = 99 \] \[ A_{47} = 49 + 47(2) = 49 + 94 = 143 \] \[ A_{49} = 49 + 49(2) = 49 + 98 = 147 \] 4. Calculate the Mean: \[ \text{Mean} = \frac{51 + 99 + 143 + 147}{4} \] \[ \text{Sum} = 51 + 99 = 150 \] \[ \text{Sum} = 143 + 147 = 290 \] \[ \text{Mean} = \frac{150 + 290}{4} = \frac{440}{4} = 110 \] (Note: Let's re-verify the terms. If the question implies different indices or boundary values, common values like 130 or 120 occur. Based on current values, mean is 110).
Step 4: Final Answer:
The mean of the specified terms is 110.