Let `A_(1), A_(2), A_(3),….., A_(n)` be squares such that for each `n ge 1` the length of a side of `A _(n)` equals the length of a diagonal of `A _(n+1).` If the side of `A_(1)` be 20 units then the smallest value of 'n' for which area of `A_(n)` is less than 1.

Options

A. 7
B. 8
C. 9
D. 10

Answer

To solve the problem, we need to find the smallest value of \( n \) such that the area of square \( A_n \) is less than 1, given that the side length of \( A_1 \) is 20 units and the side length of \( A_n \) is defined in terms of the diagonal of the next square \( A_{n+1} \).

### Step-by-step Solution:

1. **Define the side lengths**:
Let the side length of square \( A_n \) be denoted as \( a_n \). We know that:
\[ a_1 = 20 \text{ units} \]

2. **Relationship between side lengths**:
The diagonal \( d \) of a square with side length \( a \) is given by the formula:
\[ d = a \sqrt{2} \]
According to the problem, the side length of square \( A_n \) is equal to the diagonal of square \( A_{n+1} \):
\[ a_n = a_{n+1} \sqrt{2} \]

3. **Expressing side lengths recursively**:
Rearranging the equation gives:
\[ a_{n+1} = \frac{a_n}{\sqrt{2}} \]
This means that each subsequent side length is \( \frac{1}{\sqrt{2}} \) times the previous side length.

4. **Finding a general formula**:
Starting from \( a_1 = 20 \):
\[ a_2 = \frac{20}{\sqrt{2}} = 20 \cdot 2^{-1/2} = 20 \cdot 2^{-0.5} \]
\[ a_3 = \frac{a_2}{\sqrt{2}} = \frac{20 \cdot 2^{-0.5}}{\sqrt{2}} = 20 \cdot 2^{-1} = 20 \cdot 2^{-1} \]
Continuing this pattern, we can generalize:
\[ a_n = 20 \cdot 2^{-\frac{n-1}{2}} \]

5. **Calculating the area**:
The area \( A_n \) of square \( A_n \) is given by:
\[ A_n = a_n^2 = \left(20 \cdot 2^{-\frac{n-1}{2}}\right)^2 = 400 \cdot 2^{-(n-1)} \]

6. **Setting up the inequality**:
We need to find the smallest \( n \) such that:
\[ A_n < 1 \]
This translates to:
\[ 400 \cdot 2^{-(n-1)} < 1 \]
Dividing both sides by 400 gives:
\[ 2^{-(n-1)} < \frac{1}{400} \]

7. **Taking logarithms**:
Taking logarithm base 2 on both sides:
\[ -(n-1) < \log_2\left(\frac{1}{400}\right) \]
This simplifies to:
\[ n - 1 > -\log_2(400) \]
Thus:
\[ n > 1 - \log_2(400) \]

8. **Calculating \( \log_2(400) \)**:
We know \( 400 = 2^4 \cdot 25 = 2^4 \cdot 5^2 \):
\[ \log_2(400) = \log_2(2^4) + \log_2(5^2) = 4 + 2\log_2(5) \]
Using \( \log_2(5) \approx 2.32193 \):
\[ \log_2(400) \approx 4 + 2 \cdot 2.32193 \approx 8.64386 \]

9. **Finding \( n \)**:
Now, substituting back:
\[ n > 1 - 8.64386 \implies n > -7.64386 \]
Since \( n \) must be a positive integer, we check for values of \( n \) starting from 1 upwards.

10. **Testing values**:
- For \( n = 10 \):
\[ A_{10} = 400 \cdot 2^{-9} = \frac{400}{512} \approx 0.78125 < 1 \]
- For \( n = 9 \):
\[ A_{9} = 400 \cdot 2^{-8} = \frac{400}{256} \approx 1.5625 > 1 \]

Thus, the smallest value of \( n \) for which the area of \( A_n \) is less than 1 is:
\[ \boxed{10} \]