To insert 10 arithmetic means between 3 and 80, we follow these steps:
### Step 1: Identify the first and last terms
The first term (a) is 3, and the last term (l) is 80.
### Step 2: Determine the total number of terms
Since we are inserting 10 arithmetic means between 3 and 80, the total number of terms (n) will be:
\[ n = 10 + 2 = 12 \]
(10 means + 1st term + last term)
### Step 3: Use the formula for the last term of an arithmetic progression
The formula for the last term of an arithmetic progression (AP) is given by:
\[ l = a + (n - 1)d \]
where:
- \( l \) is the last term,
- \( a \) is the first term,
- \( n \) is the total number of terms,
- \( d \) is the common difference.
### Step 4: Substitute the known values into the formula
Substituting the known values into the formula:
\[ 80 = 3 + (12 - 1)d \]
\[ 80 = 3 + 11d \]
### Step 5: Solve for the common difference (d)
Rearranging the equation to solve for \( d \):
\[ 80 - 3 = 11d \]
\[ 77 = 11d \]
\[ d = \frac{77}{11} = 7 \]
### Step 6: Write the arithmetic sequence
Now that we have \( a = 3 \) and \( d = 7 \), we can write the sequence:
- First term: \( a = 3 \)
- First arithmetic mean: \( a + d = 3 + 7 = 10 \)
- Second arithmetic mean: \( a + 2d = 3 + 2 \times 7 = 17 \)
- Third arithmetic mean: \( a + 3d = 3 + 3 \times 7 = 24 \)
- Fourth arithmetic mean: \( a + 4d = 3 + 4 \times 7 = 31 \)
- Fifth arithmetic mean: \( a + 5d = 3 + 5 \times 7 = 38 \)
- Sixth arithmetic mean: \( a + 6d = 3 + 6 \times 7 = 45 \)
- Seventh arithmetic mean: \( a + 7d = 3 + 7 \times 7 = 52 \)
- Eighth arithmetic mean: \( a + 8d = 3 + 8 \times 7 = 59 \)
- Ninth arithmetic mean: \( a + 9d = 3 + 9 \times 7 = 66 \)
- Tenth arithmetic mean: \( a + 10d = 3 + 10 \times 7 = 73 \)
- Last term: \( l = 80 \)
### Step 7: List the complete sequence
The complete sequence including the 10 arithmetic means is:
\[ 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80 \]
### Final Answer
The 10 arithmetic means inserted between 3 and 80 are:
\[ 10, 17, 24, 31, 38, 45, 52, 59, 66, 73 \]
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