15. $$ S_{n} $$ of an arithmetic series is given by $$ S_{n}=n^{2}+6 n $$. Find $$ S_{1} $$ and $$ S_{2} $$ and hence write down the values of $$ T_{1} $$ and $$ T_{2} $$.

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To solve the problem, we start with the formula for the sum of the first $$ n $$ terms of an arithmetic series given by:

$$
S_{n} = n^{2} + 6n
$$

### Step 1: Calculate $$ S_{1} $$

To find $$ S_{1} $$, we substitute $$ n = 1 $$ into the formula:

$$
S_{1} = 1^{2} + 6 \cdot 1
$$

Calculating this gives:

$$
S_{1} = 1 + 6 = 7
$$

### Step 2: Calculate $$ S_{2} $$

Next, we calculate $$ S_{2} $$ by substituting $$ n = 2 $$ into the formula:

$$
S_{2} = 2^{2} + 6 \cdot 2
$$

Calculating this gives:

$$
S_{2} = 4 + 12 = 16
$$

### Step 3: Find $$ T_{1} $$ and $$ T_{2} $$

The $$ n $$-th term $$ T_{n} $$ of an arithmetic series can be found using the relationship:

$$
T_{n} = S_{n} - S_{n-1}
$$

#### Finding $$ T_{1} $$

For $$ T_{1} $$:

$$
T_{1} = S_{1} - S_{0}
$$

Since $$ S_{0} $$ is the sum of zero terms, it is defined as 0:

$$
S_{0} = 0
$$

Thus,

$$
T_{1} = S_{1} - S_{0} = 7 - 0 = 7
$$

#### Finding $$ T_{2} $$

For $$ T_{2} $$:

$$
T_{2} = S_{2} - S_{1}
$$

Substituting the values we found:

$$
T_{2} = 16 - 7 = 9
$$

### Final Results

- $$ S_{1} = 7 $$
- $$ S_{2} = 16 $$
- $$ T_{1} = 7 $$
- $$ T_{2} = 9 $$

Thus, the values are:

$$
S_{1} = 7, \quad S_{2} = 16, \quad T_{1} = 7, \quad T_{2} = 9
$$
$$ S_{1} = 7 $$, $$ S_{2} = 16 $$, $$ T_{1} = 7 $$, $$ T_{2} = 9 $$.

15. \( S_{n} \) of an arithmetic series is given by \( S_{n}=n^{2}+6 n \). Find \( S_{1} \) and \( S_{2} \) and hence write down the values of \( T_{1} \) and \( T_{2} \).

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