Three distinct non-zero real numbers form an A.P. and the squares of these numbers taken in same order form a G.P.If possible common ratio of G.P. are `3 pm sqrtn, n in N ` then `n =`____________.

Answer

To solve the problem, we start by letting the three distinct non-zero real numbers that form an arithmetic progression (A.P.) be represented as \( a - d \), \( a \), and \( a + d \), where \( d \) is the common difference.

### Step 1: Express the squares of the numbers
The squares of these numbers are:
- \( (a - d)^2 = a^2 - 2ad + d^2 \)
- \( a^2 \)
- \( (a + d)^2 = a^2 + 2ad + d^2 \)

### Step 2: Set up the condition for geometric progression (G.P.)
Since the squares of these numbers form a geometric progression, the ratio of the first term to the second term must equal the ratio of the second term to the third term:
\[ \frac{a^2}{(a - d)^2} = \frac{(a + d)^2}{a^2} \]

### Step 3: Cross-multiply
Cross-multiplying gives us:
\[ a^2 \cdot a^2 = (a - d)^2 \cdot (a + d)^2 \]
This simplifies to:
\[ a^4 = (a^2 - d^2)^2 \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ a^4 = a^4 - 2a^2d^2 + d^4 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0 = -2a^2d^2 + d^4 \]
or
\[ d^4 - 2a^2d^2 = 0 \]

### Step 6: Factor the equation
Factoring out \( d^2 \):
\[ d^2(d^2 - 2a^2) = 0 \]
This gives us two cases:
1. \( d^2 = 0 \) (not possible since the numbers must be distinct)
2. \( d^2 - 2a^2 = 0 \) which implies \( d^2 = 2a^2 \)

### Step 7: Solve for \( d \)
Taking the square root gives:
\[ d = \pm \sqrt{2} a \]

### Step 8: Substitute \( d \) back into the ratio condition
Now substituting \( d \) back into the ratio condition:
\[ \frac{a^2}{(a - d)^2} = \frac{(a + d)^2}{a^2} \]
Substituting \( d = \sqrt{2} a \) gives:
\[ \frac{a^2}{(a - \sqrt{2} a)^2} = \frac{(a + \sqrt{2} a)^2}{a^2} \]
This simplifies to:
\[ \frac{a^2}{(1 - \sqrt{2})^2 a^2} = \frac{(1 + \sqrt{2})^2 a^2}{a^2} \]
Thus:
\[ \frac{1}{(1 - \sqrt{2})^2} = (1 + \sqrt{2})^2 \]

### Step 9: Calculate the common ratio
Calculating the common ratio:
\[ (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2} \]
And:
\[ (1 - \sqrt{2})^2 = 1 - 2 + 2\sqrt{2} = 3 - 2\sqrt{2} \]

### Step 10: Relate to the given common ratio
The common ratios we found are \( 3 + 2\sqrt{2} \) and \( 3 - 2\sqrt{2} \). The problem states that the possible common ratios are \( 3 \pm \sqrt{n} \).

### Step 11: Compare and solve for \( n \)
Setting \( 2\sqrt{2} = \sqrt{n} \):
\[ n = (2\sqrt{2})^2 = 8 \]

Thus, the value of \( n \) is:
\[ \boxed{8} \]