If 11 arithmetic means are inserted between 20 and 10, the number of integral arithmetic means are

Options

A. 1
B. 6
C. 7
D. 8

Answer

To solve the problem of finding the number of integral arithmetic means inserted between 20 and 10, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Terms**:
We have the first term \( a = 10 \) and the last term \( l = 20 \). We need to insert 11 arithmetic means between these two terms.

2. **Determine the Total Number of Terms**:
The total number of terms in the arithmetic progression (AP) will be:
\[ n = 11 + 2 = 13 \]
(11 means + the first and last term).

3. **Use the Formula for the nth Term of an AP**:
The formula for the nth term of an AP is given by:
\[ a_n = a + (n-1)d \]
Here, \( n = 13 \), \( a = 10 \), and \( a_n = 20 \).

4. **Set Up the Equation**:
Substitute the known values into the formula:
\[ 20 = 10 + (13-1)d \]
Simplifying gives:
\[ 20 = 10 + 12d \]

5. **Solve for the Common Difference \( d \)**:
Rearranging the equation:
\[ 12d = 20 - 10 \]
\[ 12d = 10 \]
\[ d = \frac{10}{12} = \frac{5}{6} \]

6. **Write the Terms of the AP**:
The terms of the AP can be expressed as:
- First term: \( 10 \)
- Second term: \( 10 + d = 10 + \frac{5}{6} = \frac{65}{6} \)
- Third term: \( 10 + 2d = 10 + 2 \cdot \frac{5}{6} = \frac{70}{6} \)
- Continuing this way up to the 11th term.

The general term for the arithmetic means can be expressed as:
\[ a_k = 10 + k \cdot \frac{5}{6} \quad (k = 1, 2, \ldots, 11) \]

7. **Determine When the Terms are Integers**:
For \( a_k \) to be an integer:
\[ 10 + k \cdot \frac{5}{6} \text{ must be an integer.} \]
This implies:
\[ k \cdot \frac{5}{6} \text{ must be an integer.} \]
Therefore, \( k \) must be a multiple of 6.

8. **Count the Valid Values of \( k \)**:
The values of \( k \) that are multiples of 6 within the range \( 1 \) to \( 11 \) are:
- \( k = 6 \)

Hence, there is only **one integral arithmetic mean**.

### Final Answer:
The number of integral arithmetic means inserted between 20 and 10 is **1**.