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 break down the information given and use properties of arithmetic progressions (A.P.).

### Step 1: Understand the terms of the A.P.
In an arithmetic progression, the nth term \( T_n \) can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write down the given equation
We are given:
\[ T_1 + T_5 + T_{10} + T_{15} + T_{20} + T_{24} = 225 \]

### Step 3: Express each term in terms of \( a \) and \( d \)
Using the formula for the nth term:
- \( T_1 = a \)
- \( T_5 = a + 4d \)
- \( T_{10} = a + 9d \)
- \( T_{15} = a + 14d \)
- \( T_{20} = a + 19d \)
- \( T_{24} = a + 23d \)

### Step 4: Substitute these into the equation
Substituting these expressions into the equation gives:
\[ a + (a + 4d) + (a + 9d) + (a + 14d) + (a + 19d) + (a + 23d) = 225 \]

### Step 5: Combine like terms
Combining all the terms:
\[ 6a + (4d + 9d + 14d + 19d + 23d) = 225 \]
Calculating the sum of the coefficients of \( d \):
\[ 4 + 9 + 14 + 19 + 23 = 69 \]
Thus, we have:
\[ 6a + 69d = 225 \]

### Step 6: Simplify the equation
Dividing the entire equation by 3:
\[ 2a + 23d = 75 \quad \text{(Equation 1)} \]

### Step 7: Find the sum of the first 24 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} (a + l) \]
where \( l \) is the last term. For the first 24 terms:
\[ S_{24} = \frac{24}{2} (T_1 + T_{24}) = 12 (T_1 + T_{24}) \]

### Step 8: Express \( T_{24} \)
Using the formula for \( T_{24} \):
\[ T_{24} = a + 23d \]
Thus:
\[ T_1 + T_{24} = a + (a + 23d) = 2a + 23d \]

### Step 9: Substitute into the sum formula
Now substituting back into the sum formula:
\[ S_{24} = 12 (2a + 23d) \]

### Step 10: Use Equation 1
From Equation 1, we know \( 2a + 23d = 75 \):
\[ S_{24} = 12 \times 75 = 900 \]

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