Find 4 numbers in A.P. such that the sum of first and fourth number is 14 and the product of second and third number is 45.

Answer

To solve the problem of finding 4 numbers in an Arithmetic Progression (A.P.) such that the sum of the first and fourth numbers is 14, and the product of the second and third numbers is 45, we can follow these steps:

### Step 1: Define the terms of the A.P.
Let the four numbers in A.P. be:
- First term: \( a - 3d \)
- Second term: \( a - d \)
- Third term: \( a + d \)
- Fourth term: \( a + 3d \)

### Step 2: Set up the equations based on the problem statement.
From the problem, we have two conditions:
1. The sum of the first and fourth numbers is 14:
\[ (a - 3d) + (a + 3d) = 14 \]
Simplifying this gives:
\[ 2a = 14 \implies a = 7 \]

2. The product of the second and third numbers is 45:
\[ (a - d)(a + d) = 45 \]
This can be rewritten using the difference of squares:
\[ a^2 - d^2 = 45 \]

### Step 3: Substitute the value of \( a \) into the second equation.
Substituting \( a = 7 \) into the equation \( a^2 - d^2 = 45 \):
\[ 7^2 - d^2 = 45 \]
This simplifies to:
\[ 49 - d^2 = 45 \]
Rearranging gives:
\[ d^2 = 49 - 45 = 4 \implies d = 2 \text{ or } d = -2 \]

### Step 4: Find the four numbers.
Now we can find the four numbers using \( a = 7 \) and \( d = 2 \):
1. First term: \( a - 3d = 7 - 3(2) = 7 - 6 = 1 \)
2. Second term: \( a - d = 7 - 2 = 5 \)
3. Third term: \( a + d = 7 + 2 = 9 \)
4. Fourth term: \( a + 3d = 7 + 3(2) = 7 + 6 = 13 \)

Thus, the four numbers in A.P. are:
\[ 1, 5, 9, 13 \]

### Step 5: Verify the conditions.
1. The sum of the first and fourth numbers:
\[ 1 + 13 = 14 \quad \text{(Condition satisfied)} \]
2. The product of the second and third numbers:
\[ 5 \times 9 = 45 \quad \text{(Condition satisfied)} \]

### Final Answer:
The four numbers in A.P. are \( 1, 5, 9, 13 \).

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