If `a_(1)+a_(5)+a_(10)+a_(15)+a_(24)=225`, then the sum of the first 24 terms of the arithmetic progression `a_(1), a_(2), a_(3)……..` is equal to

Options

A. 450
B. 675
C. 900
D. 1200

Answer

To solve the problem, we need to find the sum of the first 24 terms of the arithmetic progression given that \( a_1 + a_5 + a_{10} + a_{15} + a_{24} = 225 \).

### Step 1: Express the terms in terms of \( a_1 \) and \( d \)
In an arithmetic progression (AP), the \( n \)-th term can be expressed as:
\[ a_n = a_1 + (n-1)d \]
Thus, we can write:
- \( a_5 = a_1 + 4d \)
- \( a_{10} = a_1 + 9d \)
- \( a_{15} = a_1 + 14d \)
- \( a_{24} = a_1 + 23d \)

### Step 2: Substitute these expressions into the equation
Now substituting these expressions into the given equation:
\[ a_1 + (a_1 + 4d) + (a_1 + 9d) + (a_1 + 14d) + (a_1 + 23d) = 225 \]
This simplifies to:
\[ 5a_1 + (4d + 9d + 14d + 23d) = 225 \]
Calculating the coefficients of \( d \):
\[ 4 + 9 + 14 + 23 = 50 \]
So the equation becomes:
\[ 5a_1 + 50d = 225 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing everything by 5:
\[ a_1 + 10d = 45 \]
This gives us our first equation:
\[ a_1 + 10d = 45 \quad \text{(1)} \]

### Step 4: Find the sum of the first 24 terms
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} (a + l) \]
where \( a \) is the first term and \( l \) is the last term. For the first 24 terms:
\[ S_{24} = \frac{24}{2} (a_1 + a_{24}) = 12 (a_1 + a_{24}) \]
We need to express \( a_{24} \):
\[ a_{24} = a_1 + 23d \]
Thus:
\[ S_{24} = 12 (a_1 + (a_1 + 23d)) = 12 (2a_1 + 23d) \]

### Step 5: Substitute \( a_1 + 10d \)
From equation (1), we can express \( 23d \) in terms of \( a_1 \):
\[ d = \frac{45 - a_1}{10} \]
Substituting \( d \) back into the expression for \( S_{24} \):
\[ 23d = 23 \left(\frac{45 - a_1}{10}\right) = \frac{23(45 - a_1)}{10} \]
So:
\[ S_{24} = 12 \left(2a_1 + \frac{23(45 - a_1)}{10}\right) \]
Now we can simplify this:
\[ S_{24} = 12 \left(2a_1 + \frac{1035 - 23a_1}{10}\right) \]
Combining terms:
\[ = 12 \left(\frac{20a_1 + 1035 - 23a_1}{10}\right) = 12 \left(\frac{-3a_1 + 1035}{10}\right) \]
\[ = \frac{12(-3a_1 + 1035)}{10} = \frac{-36a_1 + 12420}{10} = -3.6a_1 + 1242 \]

### Step 6: Find the value of \( S_{24} \)
We can substitute \( a_1 \) from equation (1) into this expression, but we notice that we can also find \( S_{24} \) directly from the earlier equation:
Using \( a_1 + 10d = 45 \):
\[ S_{24} = 12(2 \cdot 45) = 12 \cdot 90 = 1080 \]

### Final Answer
Thus, the sum of the first 24 terms of the arithmetic progression is:
\[ \boxed{1080} \]