Average of 7 consecutive numbers is A. If 2 numbers which are before these 7 numbers are also taken, then what will be the new average?

Options

A. A - 1
B. A - 0.5
C. A + 1.5
D. A

Answer

To solve the problem step by step, we will first understand the average of 7 consecutive numbers and then find out the new average when 2 additional numbers are included.

### Step-by-Step Solution:

1. **Understanding the Average of 7 Consecutive Numbers:**
Let’s denote the 7 consecutive numbers as \( n, n+1, n+2, n+3, n+4, n+5, n+6 \).
The average \( A \) of these numbers can be calculated as:
\[ A = \frac{n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{7} \]
Simplifying the numerator:
\[ A = \frac{7n + (0 + 1 + 2 + 3 + 4 + 5 + 6)}{7} = \frac{7n + 21}{7} = n + 3 \]

2. **Identifying the New Average with 2 Additional Numbers:**
Now, we include 2 numbers before these 7 consecutive numbers. Let’s denote these two numbers as \( n-2 \) and \( n-1 \).
The new set of numbers will be \( n-2, n-1, n, n+1, n+2, n+3, n+4, n+5, n+6 \).

3. **Calculating the New Average:**
The new average can be calculated as:
\[ \text{New Average} = \frac{(n-2) + (n-1) + n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{9} \]
Simplifying the numerator:
\[ = \frac{(n-2) + (n-1) + n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{9} \]
\[ = \frac{9n + (0 + 1 + 2 + 3 + 4 + 5 + 6 - 2 - 1)}{9} \]
\[ = \frac{9n + 21 - 3}{9} = \frac{9n + 18}{9} = n + 2 \]

4. **Relating the New Average to A:**
Since we found that \( A = n + 3 \), we can express \( n \) in terms of \( A \):
\[ n = A - 3 \]
Therefore, the new average becomes:
\[ \text{New Average} = n + 2 = (A - 3) + 2 = A - 1 \]

### Conclusion:
The new average when 2 numbers before the 7 consecutive numbers are included is \( A - 1 \).