If sum of four numbers in A.P. is 28 and product of two middle terms is 45, then product of the first and last terms is

Options

A. 11
B. 13
C. 15
D. 17

Answer

To solve the problem step by step, we will denote the four numbers in arithmetic progression (A.P.) as follows:

Let the four numbers be:
1. \( a - 3d \)
2. \( a - d \)
3. \( a + d \)
4. \( a + 3d \)

### Step 1: Set up the equations
From the problem, we know:
- The sum of the four numbers is 28.
- The product of the two middle terms is 45.

The sum of the four numbers can be expressed as:
\[ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 4a \]
Setting this equal to 28 gives us the equation:
\[ 4a = 28 \]

### Step 2: Solve for \( a \)
Dividing both sides by 4:
\[ a = \frac{28}{4} = 7 \]

### Step 3: Use the product of the middle terms
The product of the two middle terms can be expressed as:
\[ (a - d)(a + d) = a^2 - d^2 \]
Setting this equal to 45 gives us the equation:
\[ a^2 - d^2 = 45 \]

### Step 4: Substitute \( a \) into the equation
Now, substitute \( a = 7 \) into the equation:
\[ 7^2 - d^2 = 45 \]
This simplifies to:
\[ 49 - d^2 = 45 \]

### Step 5: Solve for \( d^2 \)
Rearranging gives:
\[ d^2 = 49 - 45 = 4 \]

### Step 6: Solve for \( d \)
Taking the square root of both sides:
\[ d = \pm 2 \]

### Step 7: Find the terms of the A.P.
Now we can find the four terms of the A.P. for both values of \( d \).

1. If \( d = 2 \):
- First term: \( a - 3d = 7 - 3(2) = 1 \)
- Second term: \( a - d = 7 - 2 = 5 \)
- Third term: \( a + d = 7 + 2 = 9 \)
- Fourth term: \( a + 3d = 7 + 3(2) = 13 \)

The terms are: \( 1, 5, 9, 13 \)

2. If \( d = -2 \):
- First term: \( a - 3d = 7 - 3(-2) = 13 \)
- Second term: \( a - d = 7 - (-2) = 9 \)
- Third term: \( a + d = 7 + (-2) = 5 \)
- Fourth term: \( a + 3d = 7 + 3(-2) = 1 \)

The terms are: \( 13, 9, 5, 1 \)

### Step 8: Find the product of the first and last terms
In both cases, the product of the first and last terms is:
\[ 1 \times 13 = 13 \]
or
\[ 13 \times 1 = 13 \]

### Final Answer
Thus, the product of the first and last terms is:
\[ \boxed{13} \]