Find the sum of first 24 terms of the A.P. `a_(1) , a_(2), a_(3)`...., if it is know that `a_(1) + a_(5) + a_(10) + a_(15) + a_(20) + a_(24) = 225`

Answer

To find the sum of the first 24 terms of the arithmetic progression (A.P.) given that \( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \), we can follow these steps:

### Step 1: Express the terms of the A.P.
The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term and \( d \) is the common difference.

### Step 2: Write down the terms given in the equation.
We need to find \( a_1, a_5, a_{10}, a_{15}, a_{20}, \) and \( a_{24} \):
- \( a_1 = a_1 \)
- \( a_5 = a_1 + 4d \)
- \( a_{10} = a_1 + 9d \)
- \( a_{15} = a_1 + 14d \)
- \( a_{20} = a_1 + 19d \)
- \( a_{24} = a_1 + 23d \)

### Step 3: Set up the equation.
Now, substituting these into the given equation:
\[ a_1 + (a_1 + 4d) + (a_1 + 9d) + (a_1 + 14d) + (a_1 + 19d) + (a_1 + 23d) = 225 \]
This simplifies to:
\[ 6a_1 + (4 + 9 + 14 + 19 + 23)d = 225 \]

### Step 4: Calculate the sum of coefficients of \( d \).
Calculating the sum of the coefficients of \( d \):
\[ 4 + 9 + 14 + 19 + 23 = 69 \]
Thus, we have:
\[ 6a_1 + 69d = 225 \]

### Step 5: Rearranging the equation.
Now, we can rearrange this to express \( 6a_1 + 69d \):
\[ 6a_1 + 69d = 225 \quad \text{(Equation 1)} \]

### Step 6: Find the sum of the first 24 terms.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \]
For \( n = 24 \):
\[ S_{24} = \frac{24}{2} \times (2a_1 + 23d) = 12 \times (2a_1 + 23d) \]

### Step 7: Express \( 2a_1 + 23d \) in terms of Equation 1.
From Equation 1, we can express \( 2a_1 + 23d \):
\[ 6a_1 + 69d = 225 \]
Dividing the entire equation by 3:
\[ 2a_1 + 23d = 75 \]

### Step 8: Substitute back into the sum formula.
Now substituting this back into the sum formula:
\[ S_{24} = 12 \times 75 = 900 \]

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