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 common difference \( d \) is \( -2 \) (since each term decreases by 2).

### Step 2: Determine when the terms reach zero
To find the maximum sum, we need to determine how many terms are there until the series reaches zero. The general term \( T_n \) of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
Setting \( T_n = 0 \):
\[ 0 = 40 + (n-1)(-2) \]
\[ 0 = 40 - 2(n-1) \]
\[ 2(n-1) = 40 \]
\[ n-1 = 20 \]
\[ n = 21 \]

### Step 3: Calculate the sum of the first \( n \) terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Substituting \( n = 21 \), \( a = 40 \), and \( d = -2 \):
\[ S_{21} = \frac{21}{2} \times (2 \times 40 + (21-1)(-2)) \]
\[ = \frac{21}{2} \times (80 + 20 \times -2) \]
\[ = \frac{21}{2} \times (80 - 40) \]
\[ = \frac{21}{2} \times 40 \]
\[ = 21 \times 20 \]
\[ = 420 \]

### Conclusion
The maximum sum of the A.P. \( 40 + 38 + 36 + 34 + 32 + \ldots \) is \( 420 \).

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