Find the sum of:
`101 + 99 +97+ .... 47.`

Answer

To find the sum of the series \( 101 + 99 + 97 + \ldots + 47 \), we can follow these steps:

### Step 1: Identify the first term and the common difference
The first term \( a \) of the series is \( 101 \) and the common difference \( d \) can be calculated as:
\[ d = 99 - 101 = -2 \]

### Step 2: Determine the last term
The last term \( l \) of the series is \( 47 \).

### Step 3: Use the formula for the nth term of an arithmetic sequence
The formula for the nth term of an arithmetic sequence is given by:
\[ T_n = a + (n - 1) \cdot d \]
Setting \( T_n = 47 \) (the last term), we can substitute the values we know:
\[ 47 = 101 + (n - 1)(-2) \]

### Step 4: Solve for \( n \)
Rearranging the equation:
\[ 47 = 101 - 2(n - 1) \]
\[ 47 = 101 - 2n + 2 \]
\[ 47 = 103 - 2n \]
Now, isolate \( n \):
\[ 2n = 103 - 47 \]
\[ 2n = 56 \]
\[ n = \frac{56}{2} = 28 \]

### Step 5: Calculate the sum of the series
The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic series is:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting the values we have:
\[ S_{28} = \frac{28}{2} \cdot (101 + 47) \]
\[ S_{28} = 14 \cdot 148 \]
Now, calculate \( 14 \cdot 148 \):
\[ S_{28} = 14 \cdot 148 = 2072 \]

### Final Answer
The sum of the series \( 101 + 99 + 97 + \ldots + 47 \) is \( \boxed{2072} \).