`A , r = 1 , 2, 3 ….., n ` are n points on the parabola `y^(2)=4x` in the first quadrant . If `A_(r) = (x_(r),y_(r))` where `x_(1),x_(2),….x_(n)` are in G.P and `x_(1)=1,x_(2)=2 ` then `y_(n)` is equal to

Options

A. `2^((n+1)/2)`
B. `2^(n+1)`
C. `(sqrt(2))^(n+1)`
D. `2^(n//2)`

Answer

To solve the problem, we need to find the value of \( y_n \) for the points \( A_r = (x_r, y_r) \) on the parabola \( y^2 = 4x \) where \( x_1, x_2, \ldots, x_n \) are in geometric progression (G.P.) with \( x_1 = 1 \) and \( x_2 = 2 \).

### Step-by-Step Solution:

1. **Identify the Common Ratio of the G.P.**:
- Given \( x_1 = 1 \) and \( x_2 = 2 \), the common ratio \( r \) can be calculated as:
\[ r = \frac{x_2}{x_1} = \frac{2}{1} = 2 \]

2. **General Formula for the n-th Term of a G.P.**:
- The n-th term \( x_n \) of a G.P. can be expressed as:
\[ x_n = x_1 \cdot r^{n-1} \]
- Substituting the values we have:
\[ x_n = 1 \cdot 2^{n-1} = 2^{n-1} \]

3. **Substituting \( x_n \) into the Parabola Equation**:
- The points lie on the parabola given by \( y^2 = 4x \). We substitute \( x_n \) into this equation to find \( y_n \):
\[ y_n^2 = 4x_n = 4(2^{n-1}) = 4 \cdot 2^{n-1} \]

4. **Simplifying the Expression for \( y_n^2 \)**:
- We can express \( 4 \) as \( 2^2 \):
\[ y_n^2 = 2^2 \cdot 2^{n-1} = 2^{n+1} \]

5. **Finding \( y_n \)**:
- To find \( y_n \), we take the square root of both sides:
\[ y_n = \sqrt{y_n^2} = \sqrt{2^{n+1}} = 2^{(n+1)/2} \]

### Final Answer:
Thus, the value of \( y_n \) is:
\[ y_n = 2^{(n+1)/2} \]