If `t_(3)=15,S_(10)=120`, then the tenth term of the series is:

Answer

To solve the problem, we need to find the tenth term of an arithmetic progression (A.P.) given that the third term \( T_3 = 15 \) and the sum of the first ten terms \( S_{10} = 120 \).

### Step-by-step Solution:

1. **Identify the formulas**:
- The \( n \)-th term of an A.P. is given by:
\[ T_n = a + (n-1)d \]
- The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

2. **Set up the equations**:
- For the third term \( T_3 \):
\[ T_3 = a + 2d = 15 \quad \text{(Equation 1)} \]
- For the sum of the first ten terms \( S_{10} \):
\[ S_{10} = \frac{10}{2} \left(2a + 9d\right) = 120 \]
Simplifying this gives:
\[ 5(2a + 9d) = 120 \implies 2a + 9d = 24 \quad \text{(Equation 2)} \]

3. **Solve the equations**:
- From Equation 1:
\[ a + 2d = 15 \]
- From Equation 2:
\[ 2a + 9d = 24 \]
- We can multiply Equation 1 by 2:
\[ 2a + 4d = 30 \quad \text{(Equation 3)} \]
- Now we have:
- \( 2a + 4d = 30 \) (Equation 3)
- \( 2a + 9d = 24 \) (Equation 2)
- Subtract Equation 2 from Equation 3:
\[ (2a + 4d) - (2a + 9d) = 30 - 24 \]
This simplifies to:
\[ -5d = 6 \implies d = -\frac{6}{5} \]

4. **Find \( a \)**:
- Substitute \( d \) back into Equation 1:
\[ a + 2\left(-\frac{6}{5}\right) = 15 \]
Simplifying gives:
\[ a - \frac{12}{5} = 15 \implies a = 15 + \frac{12}{5} \]
Converting 15 to a fraction:
\[ a = \frac{75}{5} + \frac{12}{5} = \frac{87}{5} \]

5. **Find the tenth term \( T_{10} \)**:
- Using the formula for the \( n \)-th term:
\[ T_{10} = a + (10-1)d = \frac{87}{5} + 9\left(-\frac{6}{5}\right) \]
Simplifying gives:
\[ T_{10} = \frac{87}{5} - \frac{54}{5} = \frac{33}{5} \]

### Final Answer:
The tenth term of the series is:
\[ T_{10} = \frac{33}{5} \]