If `S_(n)` denotes sum of first `n` tem of a series and if `S_(n+2)-S_(n)=n^(2)` then `S_(20)` is `("given that" T_(1)+T_(2)=0)`

Options

A. `1540`
B. `1140`
C. `770`
D. `1120`

Explanation

To solve the problem, we need to find the value of \( S_{20} \) given the condition \( S_{n+2} - S_n = n^2 \) and the information that \( T_1 + T_2 = 0 \).

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
We have the equation \( S_{n+2} - S_n = n^2 \). This means the difference between the sum of the first \( n+2 \) terms and the sum of the first \( n \) terms equals \( n^2 \).

2. **Setting Up the Series:**
Let’s denote the first term of the series as \( a \) and the common difference as \( d \). The terms of the series can be expressed as:
- \( T_1 = a \)
- \( T_2 = a + d \)
- \( T_3 = a + 2d \)
- \( T_4 = a + 3d \)
- and so on.

3. **Finding \( S_n \):**
The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

4. **Using the Given Condition:**
Substitute \( n \) with \( n \) and \( n+2 \):
\[ S_{n+2} = \frac{n+2}{2} \left( 2a + (n+1)d \right) \]
Now, using the condition:
\[ S_{n+2} - S_n = n^2 \]

5. **Expanding the Sums:**
\[ \frac{n+2}{2} \left( 2a + (n+1)d \right) - \frac{n}{2} \left( 2a + (n-1)d \right) = n^2 \]
Simplifying this equation will help us find relationships between \( a \) and \( d \).

6. **Substituting Values:**
Let's put \( n = 1 \):
\[ S_3 - S_1 = 1^2 \]
This gives us:
\[ S_3 - S_1 = 1 \]
Now, we can express \( S_3 \) and \( S_1 \):
\[ S_3 = a + (a + d) + (a + 2d) = 3a + 3d \]
\[ S_1 = a \]
Thus:
\[ 3a + 3d - a = 1 \implies 2a + 3d = 1 \quad \text{(Equation 1)} \]

7. **Using the Condition \( T_1 + T_2 = 0 \):**
From \( T_1 + T_2 = 0 \):
\[ a + (a + d) = 0 \implies 2a + d = 0 \quad \text{(Equation 2)} \]

8. **Solving the Equations:**
Now, we can solve Equations 1 and 2 simultaneously:
From Equation 2, we have \( d = -2a \). Substitute this into Equation 1:
\[ 2a + 3(-2a) = 1 \implies 2a - 6a = 1 \implies -4a = 1 \implies a = -\frac{1}{4} \]
Then, substituting \( a \) back into Equation 2:
\[ d = -2(-\frac{1}{4}) = \frac{1}{2} \]

9. **Finding \( S_{20} \):**
Now we can calculate \( S_{20} \):
\[ S_{20} = \frac{20}{2} \left( 2a + (20-1)d \right) = 10 \left( 2(-\frac{1}{4}) + 19(\frac{1}{2}) \right) \]
\[ = 10 \left( -\frac{1}{2} + \frac{19}{2} \right) = 10 \left( \frac{18}{2} \right) = 10 \times 9 = 90 \]

### Final Answer:
Thus, \( S_{20} = 90 \).