In an increasing sequence of four positive integers, the first 3 terms are in A.P., the last 3 terms are in G.P. and the fourth term exceed the first term by 30, then the common difference of A.P. lying in interval `[1,9]` is:

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the solution logically.

### Step 1: Define the terms of the sequence
Let the first three terms of the sequence be:
- First term: \( a \)
- Second term: \( a + d \)
- Third term: \( a + 2d \)

The fourth term exceeds the first term by 30, so:
- Fourth term: \( a + 30 \)

### Step 2: Set up the conditions for the sequence
According to the problem:
- The first three terms \( a, a + d, a + 2d \) are in Arithmetic Progression (AP).
- The last three terms \( a + d, a + 2d, a + 30 \) are in Geometric Progression (GP).

### Step 3: Use the properties of AP and GP
For the first three terms to be in AP:
- The condition is satisfied as they are defined as \( a, a + d, a + 2d \).

For the last three terms to be in GP:
- The condition for GP is given by:
\[ (a + 2d)^2 = (a + d)(a + 30) \]

### Step 4: Expand the GP condition
Expanding the GP condition:
\[ (a + 2d)^2 = (a + d)(a + 30) \]
\[ a^2 + 4ad + 4d^2 = a^2 + 30a + ad + 30d \]

### Step 5: Simplify the equation
Now, we can simplify the equation by subtracting \( a^2 \) from both sides:
\[ 4ad + 4d^2 = 30a + ad + 30d \]
Rearranging gives:
\[ 4d^2 + 4ad - ad - 30d - 30a = 0 \]
\[ 4d^2 + 3ad - 30a - 30d = 0 \]

### Step 6: Factor the equation
We can factor the equation:
\[ 4d^2 + (3a - 30)d - 30a = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4 \), \( b = 3a - 30 \), and \( c = -30a \).

### Step 8: Find the discriminant
The discriminant \( D \) must be non-negative for \( d \) to have real solutions:
\[ D = (3a - 30)^2 - 4 \cdot 4 \cdot (-30a) \]
\[ D = (3a - 30)^2 + 480a \]

### Step 9: Set the conditions for \( d \)
Since \( d \) must be positive and lie in the interval \( [1, 9] \):
1. From \( 2d - 15 > 0 \) we get \( d > 7.5 \).
2. From \( 10 - d > 0 \) we get \( d < 10 \).

### Step 10: Determine possible values of \( d \)
Thus, the possible integer values of \( d \) are:
- \( d = 8 \)
- \( d = 9 \)

### Final Answer
The common difference \( d \) of the AP lying in the interval \( [1, 9] \) is:
- \( d = 8 \) and \( d = 9 \).