If `a_(1),a_(2),a_(3),………. `are in A.P. such that `a_(1) + a_(5) + a_(10) + a_(15) + a_(20) + a_(24) = 225, ` then `a_(1) + a_(2) + a_(3) + …… a_(23) + a_(24)` =

Options

A. 600
B. 900
C. 1200
D. 1800

Answer

To solve the problem, we start with the information given about the arithmetic progression (A.P.) and the sum of specific terms.

### Step-by-Step Solution:

1. **Understanding the A.P.**:
Let the first term of the A.P. be \( a_1 \) and the common difference be \( d \). The terms can be expressed as:
- \( a_1 = a_1 \)
- \( a_2 = a_1 + d \)
- \( a_3 = a_1 + 2d \)
- \( a_4 = a_1 + 3d \)
- \( a_5 = a_1 + 4d \)
- \( a_{10} = a_1 + 9d \)
- \( a_{15} = a_1 + 14d \)
- \( a_{20} = a_1 + 19d \)
- \( a_{24} = a_1 + 23d \)

2. **Setting up the equation**:
From the problem, we know:
\[ a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \]
Substituting the expressions for these terms:
\[ a_1 + (a_1 + 4d) + (a_1 + 9d) + (a_1 + 14d) + (a_1 + 19d) + (a_1 + 23d) = 225 \]

3. **Combining like terms**:
Combine the terms:
\[ 6a_1 + (4d + 9d + 14d + 19d + 23d) = 225 \]
Calculate the sum of the coefficients of \( d \):
\[ 4 + 9 + 14 + 19 + 23 = 69 \]
Thus, we have:
\[ 6a_1 + 69d = 225 \]

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

5. **Expressing \( 2a_1 + 23d \)**:
We can express \( 2a_1 + 23d \) in terms of \( a_1 \) and \( d \). From the equation \( 6a_1 + 69d = 225 \), we can isolate \( 2a_1 + 23d \):
\[ 2a_1 + 23d = \frac{1}{3}(6a_1 + 69d) = \frac{1}{3}(225) = 75 \]

6. **Calculating the total sum**:
Now substituting back into the sum formula:
\[ S_{24} = 12 \times (2a_1 + 23d) = 12 \times 75 = 900 \]

### Final Answer:
\[ \boxed{900} \]