Insert AMs between 7 and 71 such that `5^(th)` AM is 27. Also find the number of AMs

Explanation

To solve the problem of inserting arithmetic means (AMs) between 7 and 71 such that the 5th AM is 27, we can follow these steps:

### Step 1: Understand the Problem
We need to insert AMs between the numbers 7 and 71. The 5th AM is given as 27. The first term (a) is 7, and the last term (l) is 71.

### Step 2: Identify the Terms
The sequence of terms can be represented as:
- First term (a) = 7
- Second term (a1)
- Third term (a2)
- Fourth term (a3)
- Fifth term (a4) = 27 (given)
- Sixth term (a5) = 71

### Step 3: Use the Formula for the n-th Term of an AP
The n-th term of an arithmetic progression (AP) can be expressed as:
\[ a_n = a + (n-1) \cdot d \]
Where:
- \( a \) = first term
- \( d \) = common difference
- \( n \) = term number

### Step 4: Set Up the Equation for the 5th Term
Since the 5th AM (which is the 6th term of the sequence) is given as 27, we can write:
\[ a + 5d = 27 \]
Substituting \( a = 7 \):
\[ 7 + 5d = 27 \]

### Step 5: Solve for d
Rearranging the equation:
\[ 5d = 27 - 7 \]
\[ 5d = 20 \]
\[ d = \frac{20}{5} = 4 \]

### Step 6: Find the Number of Terms
Now we need to find the total number of terms in the sequence from 7 to 71. The last term can be expressed as:
\[ l = a + (n-1) \cdot d \]
Substituting \( l = 71 \), \( a = 7 \), and \( d = 4 \):
\[ 71 = 7 + (n-1) \cdot 4 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 71 - 7 = (n-1) \cdot 4 \]
\[ 64 = (n-1) \cdot 4 \]
\[ n-1 = \frac{64}{4} = 16 \]
\[ n = 16 + 1 = 17 \]

### Step 8: Find the Number of AMs
Since we have 17 terms in total (including the first and last terms), the number of AMs inserted between 7 and 71 is:
\[ \text{Number of AMs} = n - 2 = 17 - 2 = 15 \]

### Final Answer
Thus, the number of arithmetic means (AMs) that can be inserted between 7 and 71 such that the 5th AM is 27 is **15**.

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