img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_ALG_DPP_4_2_E01_231_Q01.png" width="80%"gt
Let `ABCD` is a unit square and each side of the square is divided in the ratio `alpha : (1-alpha) (0 lt alpha lt 1)` . These points are connected to obtain another square. The sides of new square are divided in the ratio `alpha : (1-alpha)` and points are joined to obtain another square. The process is continued idefinitely. Let `a_(n)` denote the length of side and `A_(n)` the area of the `n^(th)` square
If`alpha=(1)/(3)`, then the least value of `n` for which `A_(n) gt (1)/(10)` is

Options

A. `4`
B. `5`
C. `6`
D. `7`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a unit square \(ABCD\) with each side divided in the ratio \(\alpha : (1 - \alpha)\). We need to find the least value of \(n\) such that the area \(A_n\) of the \(n^{th}\) square is greater than \(\frac{1}{10}\).

### Step 2: Define Variables
Let:
- \(A_n\) = Area of the \(n^{th}\) square
- \(a_n\) = Side length of the \(n^{th}\) square

Since the area of a square is given by the square of its side length, we have:
\[ A_n = a_n^2 \]

### Step 3: Establish the Relationship Between Squares
From the problem, we know that the side length of the new square can be expressed in terms of the previous square's side length. The relationship is given by:
\[ a_{n+1}^2 = (1 - \alpha)^2 + \alpha^2 a_n^2 \]
Substituting \(A_n\) into the equation gives:
\[ A_{n+1} = (1 - \alpha)^2 + \alpha^2 A_n \]

### Step 4: Substitute \(\alpha = \frac{1}{3}\)
Now, substituting \(\alpha = \frac{1}{3}\):
\[ A_{n+1} = \left(1 - \frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 A_n \]
Calculating the terms:
\[ A_{n+1} = \left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2 A_n = \frac{4}{9} + \frac{1}{9} A_n \]

### Step 5: Express in Terms of \(A_n\)
Rearranging gives:
\[ A_{n+1} = \frac{4}{9} + \frac{1}{9} A_n \]

### Step 6: Find the Recurrence Relation
We can rewrite the recurrence relation as:
\[ A_{n+1} = \frac{1}{9} A_n + \frac{4}{9} \]

### Step 7: Initial Condition
Since the area of the first square \(A_1 = 1\) (as it is a unit square):
\[ A_1 = 1 \]

### Step 8: Calculate Subsequent Areas
Using the recurrence relation:
- For \(n = 1\):
\[ A_2 = \frac{1}{9} \cdot 1 + \frac{4}{9} = \frac{5}{9} \]
- For \(n = 2\):
\[ A_3 = \frac{1}{9} \cdot \frac{5}{9} + \frac{4}{9} = \frac{5}{81} + \frac{36}{81} = \frac{41}{81} \]
- For \(n = 3\):
\[ A_4 = \frac{1}{9} \cdot \frac{41}{81} + \frac{4}{9} = \frac{41}{729} + \frac{324}{729} = \frac{365}{729} \]
- For \(n = 4\):
\[ A_5 = \frac{1}{9} \cdot \frac{365}{729} + \frac{4}{9} = \frac{365}{6561} + \frac{2916}{6561} = \frac{3281}{6561} \]

### Step 9: Check When \(A_n > \frac{1}{10}\)
We need to find the smallest \(n\) such that:
\[ A_n > \frac{1}{10} \]

Calculating the values:
- \(A_1 = 1 > \frac{1}{10}\)
- \(A_2 = \frac{5}{9} \approx 0.555 > \frac{1}{10}\)
- \(A_3 = \frac{41}{81} \approx 0.506 > \frac{1}{10}\)
- \(A_4 = \frac{365}{729} \approx 0.500 > \frac{1}{10}\)
- \(A_5 = \frac{3281}{6561} \approx 0.500 > \frac{1}{10}\)

### Conclusion
The least value of \(n\) for which \(A_n > \frac{1}{10}\) is \(n = 1\).