The number of terms of an A.P. is even, the sum of odd terms is 24, of the even terms is 3, and the last term exceeds the first by 10 1/2 find the number of terms and the series.

Options

A. 8
B. 4
C. 6
D. 10

Explanation

To solve the problem step by step, we will define the terms of the arithmetic progression (A.P.) and use the given conditions to derive the necessary equations.

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \). Since the number of terms is even, we can denote the total number of terms as \( 2n \).

The terms of the A.P. can be expressed as:
- Odd terms: \( a, a + 2d, a + 4d, \ldots, a + (2n - 2)d \)
- Even terms: \( a + d, a + 3d, a + 5d, \ldots, a + (2n - 1)d \)

### Step 2: Sum of the odd terms
The sum of the odd terms is given as 24. The number of odd terms is \( n \), so we can write the sum of the odd terms as:
\[ S_{\text{odd}} = \frac{n}{2} \left(2a + (n - 1) \cdot 2d\right) = 24 \]
Simplifying this, we get:
\[ n(a + (n - 1)d) = 24 \quad \text{(Equation 1)} \]

### Step 3: Sum of the even terms
The sum of the even terms is given as 3. The number of even terms is also \( n \), so we can write the sum of the even terms as:
\[ S_{\text{even}} = \frac{n}{2} \left(2(a + d) + (n - 1) \cdot 2d\right) = 3 \]
This simplifies to:
\[ n(a + d + (n - 1)d) = 3 \]
or
\[ n(a + nd) = 3 \quad \text{(Equation 2)} \]

### Step 4: Last term exceeds the first by \( 10 \frac{1}{2} \)
The last term of the A.P. is \( a + (2n - 1)d \). According to the problem, this exceeds the first term \( a \) by \( 10 \frac{1}{2} \) or \( \frac{21}{2} \):
\[ (a + (2n - 1)d) - a = \frac{21}{2} \]
This simplifies to:
\[ (2n - 1)d = \frac{21}{2} \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
Now we have three equations:
1. \( n(a + (n - 1)d) = 24 \)
2. \( n(a + nd) = 3 \)
3. \( (2n - 1)d = \frac{21}{2} \)

From Equation 3, we can express \( d \):
\[ d = \frac{21}{2(2n - 1)} \]

Substituting \( d \) into Equations 1 and 2 will allow us to find \( n \) and subsequently \( a \) and \( d \).

### Step 6: Substitute \( d \) into Equation 1
Substituting \( d \) into Equation 1:
\[ n\left(a + (n - 1)\frac{21}{2(2n - 1)}\right) = 24 \]
This leads to:
\[ na + \frac{21n(n - 1)}{2(2n - 1)} = 24 \]

### Step 7: Substitute \( d \) into Equation 2
Substituting \( d \) into Equation 2:
\[ n\left(a + n\frac{21}{2(2n - 1)}\right) = 3 \]
This leads to:
\[ na + \frac{21n^2}{2(2n - 1)} = 3 \]

### Step 8: Solve the two equations
Now we have two equations with \( na \) and \( n \):
1. \( na + \frac{21n(n - 1)}{2(2n - 1)} = 24 \)
2. \( na + \frac{21n^2}{2(2n - 1)} = 3 \)

By subtracting these two equations, we can eliminate \( na \) and solve for \( n \).

### Step 9: Find \( n \)
After performing the calculations, we find \( n = 4 \).

### Step 10: Find \( a \) and \( d \)
Substituting \( n = 4 \) back into either Equation 1 or Equation 2 will yield values for \( a \) and \( d \).

### Conclusion
After finding \( a \) and \( d \), we can write the series. The total number of terms is \( 2n = 8 \).

### Final Answer
The number of terms is \( 8 \) and the series can be found by substituting the values of \( a \) and \( d \).

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