In an A.P., if `T_(1) +T_(5)+ T_(10) +T_(15)+ T_(20) + T_(24) = 225,` find the sum of its 24 terms.

Explanation

To solve the problem step by step, we will use the properties of an arithmetic progression (A.P.) and the formula for the sum of the first n terms.

### Step 1: Understand the given information
We know that:
- The sum of specific terms in an A.P. is given as:
\[ T_1 + T_5 + T_{10} + T_{15} + T_{20} + T_{24} = 225 \]

### Step 2: Write the general formula for the n-th term of an A.P.
The n-th term of an A.P. can be expressed as:
\[ T_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 3: Express the terms in the equation
Using the formula for the n-th term, we can express each term in the sum:
- \( T_1 = a \)
- \( T_5 = a + 4d \)
- \( T_{10} = a + 9d \)
- \( T_{15} = a + 14d \)
- \( T_{20} = a + 19d \)
- \( T_{24} = a + 23d \)

### Step 4: Substitute the terms into the equation
Now, substituting these into the equation:
\[ T_1 + T_5 + T_{10} + T_{15} + T_{20} + T_{24} = a + (a + 4d) + (a + 9d) + (a + 14d) + (a + 19d) + (a + 23d) \]
This simplifies to:
\[ 6a + (4d + 9d + 14d + 19d + 23d) = 225 \]
Calculating the sum of the coefficients of \( d \):
\[ 4 + 9 + 14 + 19 + 23 = 69 \]
Thus, we have:
\[ 6a + 69d = 225 \]

### Step 5: Rearrange the equation
We can rearrange this equation to isolate \( a \) and \( d \):
\[ 6a + 69d = 225 \]
Dividing the entire equation by 3:
\[ 2a + 23d = 75 \]

### Step 6: Find the sum of the first 24 terms
The formula for the sum of the first n terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For \( n = 24 \):
\[ S_{24} = \frac{24}{2} \times (2a + 23d) = 12 \times (2a + 23d) \]
Substituting \( 2a + 23d = 75 \) into the equation:
\[ S_{24} = 12 \times 75 = 900 \]

### Final Answer
The sum of the first 24 terms of the A.P. is:
\[ \boxed{900} \]