Insert 10 AM.'s between 5 and -17 and prove that their sum is ten times the A.M. between them.

Answer

To solve the problem of inserting 10 arithmetic means between 5 and -17, and proving that their sum is ten times the arithmetic mean between them, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the terms**: We have two terms, \( a_0 = 5 \) and \( a_{12} = -17 \). We need to insert 10 arithmetic means between these two terms.

2. **Determine the total number of terms**: The total number of terms in this arithmetic progression (AP) will be \( 12 \) (including the two given terms and the 10 means).

3. **Use the formula for the last term of an AP**: The formula for the \( n \)-th term of an AP is given by:
\[ a_n = a + (n - 1)d \]
Here, \( a = 5 \), \( a_{12} = -17 \), and \( n = 12 \).

4. **Set up the equation**:
\[ -17 = 5 + (12 - 1)d \]
Simplifying gives:
\[ -17 = 5 + 11d \]

5. **Solve for \( d \)**:
\[ -17 - 5 = 11d \implies -22 = 11d \implies d = -2 \]

6. **Calculate the arithmetic means**: Now we can find the 10 arithmetic means:
- \( a_1 = a_0 + d = 5 - 2 = 3 \)
- \( a_2 = a_1 + d = 3 - 2 = 1 \)
- \( a_3 = a_2 + d = 1 - 2 = -1 \)
- \( a_4 = a_3 + d = -1 - 2 = -3 \)
- \( a_5 = a_4 + d = -3 - 2 = -5 \)
- \( a_6 = a_5 + d = -5 - 2 = -7 \)
- \( a_7 = a_6 + d = -7 - 2 = -9 \)
- \( a_8 = a_7 + d = -9 - 2 = -11 \)
- \( a_9 = a_8 + d = -11 - 2 = -13 \)
- \( a_{10} = a_9 + d = -13 - 2 = -15 \)

Thus, the arithmetic means are: \( 3, 1, -1, -3, -5, -7, -9, -11, -13, -15 \).

7. **Calculate the sum of the arithmetic means**:
\[ \text{Sum} = 3 + 1 - 1 - 3 - 5 - 7 - 9 - 11 - 13 - 15 = -60 \]

8. **Calculate the arithmetic mean between 5 and -17**:
\[ \text{Arithmetic Mean} = \frac{5 + (-17)}{2} = \frac{-12}{2} = -6 \]

9. **Prove the relationship**:
The sum of the 10 arithmetic means is \( -60 \), and we need to check if this is \( 10 \) times the arithmetic mean:
\[ 10 \times (-6) = -60 \]

### Conclusion:
The sum of the 10 arithmetic means is indeed \( 10 \) times the arithmetic mean between \( 5 \) and \( -17 \).