The absolute difference of minimum and maximum sum of the series -21,-19,-17,…. And 21,19,17 ,….. Is :

Options

A. 121
B. 242
C. 221
D. 200

Answer

To solve the problem, we need to find the absolute difference between the minimum and maximum sums of the two series given:

1. The first series: -21, -19, -17, ...
2. The second series: 21, 19, 17, ...

### Step 1: Analyze the first series for minimum sum
The first series is an arithmetic series where:
- First term \( a = -21 \)
- Common difference \( d = 2 \)

To find the minimum sum, we need to consider the maximum number of negative terms. The last term \( a_n \) must be less than or equal to 0.

Using the formula for the nth term of an arithmetic sequence:
\[ a_n = a + (n - 1) \cdot d \]
Setting \( a_n < 0 \):
\[ -21 + (n - 1) \cdot 2 < 0 \]

### Step 2: Solve the inequality
Rearranging the inequality:
\[ 2(n - 1) < 21 \]
\[ n - 1 < 10.5 \]
\[ n < 11.5 \]

Thus, the maximum integer value for \( n \) is 11.

### Step 3: Calculate the last term for \( n = 11 \)
Now, we calculate the last term when \( n = 11 \):
\[ a_{11} = -21 + (11 - 1) \cdot 2 = -21 + 20 = -1 \]

### Step 4: Calculate the minimum sum
The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \cdot (a + a_n) \]
Substituting \( n = 11 \), \( a = -21 \), and \( a_{11} = -1 \):
\[ S_{11} = \frac{11}{2} \cdot (-21 - 1) = \frac{11}{2} \cdot (-22) = 11 \cdot (-11) = -121 \]

### Step 5: Analyze the second series for maximum sum
The second series is also an arithmetic series where:
- First term \( a = 21 \)
- Common difference \( d = -2 \)

To find the maximum sum, we need to consider the maximum number of positive terms. The last term \( a_n \) must be greater than or equal to 0.

Using the formula for the nth term:
\[ a_n = a + (n - 1) \cdot d \]
Setting \( a_n \geq 0 \):
\[ 21 + (n - 1)(-2) \geq 0 \]

### Step 6: Solve the inequality
Rearranging the inequality:
\[ 21 - 2(n - 1) \geq 0 \]
\[ 21 - 2n + 2 \geq 0 \]
\[ 23 \geq 2n \]
\[ n \leq 11.5 \]

Thus, the maximum integer value for \( n \) is also 11.

### Step 7: Calculate the last term for \( n = 11 \)
Now, we calculate the last term when \( n = 11 \):
\[ a_{11} = 21 + (11 - 1)(-2) = 21 - 20 = 1 \]

### Step 8: Calculate the maximum sum
Using the sum formula:
\[ S_{11} = \frac{n}{2} \cdot (a + a_n) \]
Substituting \( n = 11 \), \( a = 21 \), and \( a_{11} = 1 \):
\[ S_{11} = \frac{11}{2} \cdot (21 + 1) = \frac{11}{2} \cdot 22 = 11 \cdot 11 = 121 \]

### Step 9: Calculate the absolute difference
Now, we find the absolute difference between the minimum and maximum sums:
\[ |S_{\text{min}} - S_{\text{max}}| = |-121 - 121| = |-242| = 242 \]

### Final Answer
The absolute difference of the minimum and maximum sum of the series is:
\[ \boxed{242} \]