The number of terms of an A.P. is even; the sum of the odd terms is 310; the sum of the even terms is 340; the last term exceeds the first by 57. Then,

Options

A. (a) Number of terms of the series `= 20`
B. (b) Number of terms of the series `= 10`
C. (c) The first term of series = 4
D. (d) The first term of series = 5

Answer

To solve the problem step by step, we will break down the information given and use the properties of an arithmetic progression (A.P.).

### Step-by-Step Solution

**Step 1: Define the A.P.**
Let the first term of the A.P. be \( a_1 \) and the common difference be \( d \). The number of terms is given to be \( 2n \) (even).

**Step 2: Write the sums of odd and even terms.**
The odd terms of the A.P. are:
- \( a_1, a_3, a_5, \ldots, a_{2n-1} \)

The sum of the odd terms can be expressed as:
\[ S_{\text{odd}} = a_1 + (a_1 + 2d) + (a_1 + 4d) + \ldots + (a_1 + (2n-2)d) \]
This can be simplified to:
\[ S_{\text{odd}} = n a_1 + d(0 + 2 + 4 + \ldots + (2n-2)) \]
The sum of the first \( n-1 \) even numbers is \( (n-1)n \), so:
\[ S_{\text{odd}} = n a_1 + d(n-1)n \]
Given \( S_{\text{odd}} = 310 \), we have:
\[ n a_1 + d(n-1)n = 310 \quad \text{(Equation 1)} \]

The even terms of the A.P. are:
- \( a_2, a_4, a_6, \ldots, a_{2n} \)

The sum of the even terms can be expressed as:
\[ S_{\text{even}} = (a_1 + d) + (a_1 + 3d) + (a_1 + 5d) + \ldots + (a_1 + (2n-1)d) \]
This can be simplified to:
\[ S_{\text{even}} = n(a_1 + d) + d(0 + 2 + 4 + \ldots + (2n-2)) = n a_1 + n d + d(n-1)n \]
Given \( S_{\text{even}} = 340 \), we have:
\[ n a_1 + n d + d(n-1)n = 340 \quad \text{(Equation 2)} \]

**Step 3: Subtract Equation 1 from Equation 2.**
Subtracting Equation 1 from Equation 2 gives:
\[ (n a_1 + n d + d(n-1)n) - (n a_1 + d(n-1)n) = 340 - 310 \]
This simplifies to:
\[ n d = 30 \quad \Rightarrow \quad d = \frac{30}{n} \quad \text{(Equation 3)} \]

**Step 4: Use the information about the last term.**
The last term \( a_{2n} \) exceeds the first term \( a_1 \) by 57:
\[ a_{2n} - a_1 = 57 \]
Using the formula for the last term:
\[ a_{2n} = a_1 + (2n-1)d \]
Thus:
\[ (a_1 + (2n-1)d) - a_1 = 57 \quad \Rightarrow \quad (2n-1)d = 57 \quad \text{(Equation 4)} \]

**Step 5: Substitute \( d \) from Equation 3 into Equation 4.**
Substituting \( d = \frac{30}{n} \) into Equation 4 gives:
\[ (2n-1) \left(\frac{30}{n}\right) = 57 \]
Multiplying through by \( n \):
\[ 30(2n-1) = 57n \]
Expanding and rearranging:
\[ 60n - 30 = 57n \]
\[ 3n = 30 \quad \Rightarrow \quad n = 10 \]

**Step 6: Find the total number of terms.**
Since the number of terms is \( 2n \):
\[ \text{Total number of terms} = 2 \times 10 = 20 \]

**Step 7: Find the first term \( a_1 \).**
Substituting \( n = 10 \) back into Equation 3 to find \( d \):
\[ d = \frac{30}{10} = 3 \]
Now substituting \( n = 10 \) and \( d = 3 \) into Equation 1:
\[ 10 a_1 + 3(10-1)10 = 310 \]
\[ 10 a_1 + 3 \times 90 = 310 \]
\[ 10 a_1 + 270 = 310 \]
\[ 10 a_1 = 40 \quad \Rightarrow \quad a_1 = 4 \]

### Final Answers
- The number of terms in the A.P. is **20**.
- The first term of the A.P. is **4**.