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 1: Understand the Problem
We need to insert 11 arithmetic means between the numbers 20 and 10. This means we will have a total of 13 terms in an arithmetic progression (AP): the first term (a1) is 20, the last term (a13) is 10, and the 11 terms in between are a2, a3, ..., a12.

### Step 2: Use the Formula for the nth Term of an AP
The nth term of an arithmetic progression can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

### Step 3: Set Up the Equation
For our case:
- \( a_1 = 20 \)
- \( a_{13} = 10 \)
- We need to find \( d \).

Using the formula for the 13th term:
\[ a_{13} = a_1 + (13 - 1)d \]
Substituting the known values:
\[ 10 = 20 + 12d \]

### Step 4: Solve for d
Rearranging the equation gives:
\[ 10 - 20 = 12d \]
\[ -10 = 12d \]
\[ d = -\frac{10}{12} = -\frac{5}{6} \]

### Step 5: Write the General Term
Now, we can express the general term \( a_n \):
\[ a_n = 20 + (n-1)\left(-\frac{5}{6}\right) \]
\[ a_n = 20 - \frac{5(n-1)}{6} \]

### Step 6: Find Integral Values
For \( a_n \) to be an integer, the term \( -\frac{5(n-1)}{6} \) must also be an integer. This means that \( n-1 \) must be a multiple of 6.

Let:
\[ n - 1 = 6k \]
Then:
\[ n = 6k + 1 \]

### Step 7: Determine k Values
Since we need \( n \) to be between 2 and 12 (inclusive) to ensure we have 11 arithmetic means:
- For \( k = 0 \): \( n = 1 \) (not valid, as it corresponds to the first term)
- For \( k = 1 \): \( n = 7 \) (valid)
- For \( k = 2 \): \( n = 13 \) (not valid, exceeds the number of means)

### Conclusion
The only valid value for \( k \) is 1, which gives us \( n = 7 \). This means there is only **1 integral arithmetic mean** that can be inserted between 20 and 10.