Find the maximum sum of the A.P. `40 +38+36+34+32+……………`

Answer

To find the maximum sum of the arithmetic progression (A.P.) given by the series \( 40 + 38 + 36 + 34 + 32 + \ldots \), we can follow these steps:

### Step 1: Identify the first term and common difference
The first term \( a \) of the A.P. is \( 40 \), and the second term is \( 38 \). The common difference \( d \) can be calculated as:
\[ d = 38 - 40 = -2 \]

### Step 2: Determine the last term before the sum becomes negative
Since the series is decreasing, we need to find the last term that is non-negative. We set the nth term \( a_n \) to \( 0 \):
\[ a_n = a + (n-1) \cdot d \]
Substituting the known values:
\[ 0 = 40 + (n-1)(-2) \]
Rearranging gives:
\[ 0 = 40 - 2(n-1) \]
\[ 2(n-1) = 40 \]
\[ n-1 = 20 \]
\[ n = 21 \]
Thus, there are \( 21 \) terms in the A.P. before it becomes negative.

### Step 3: Calculate the sum of the first n terms
The formula for the sum \( S_n \) of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Substituting the values we found:
\[ S_{21} = \frac{21}{2} \cdot (2 \cdot 40 + (21-1)(-2)) \]
Calculating inside the parentheses:
\[ 2 \cdot 40 = 80 \]
\[ (21-1)(-2) = 20 \cdot (-2) = -40 \]
Thus:
\[ S_{21} = \frac{21}{2} \cdot (80 - 40) = \frac{21}{2} \cdot 40 \]
Calculating further:
\[ S_{21} = \frac{21 \cdot 40}{2} = \frac{840}{2} = 420 \]

### Final Answer
The maximum sum of the A.P. \( 40 + 38 + 36 + 34 + 32 + \ldots \) is \( \boxed{420} \).

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