A particle moves in a straight line with constant acceleration a. The displacements of particle from origin in times `t_1, t_2 and t_3 are s_1,s_2 and s_3` respectively. If times are in AP with common difference d and displacements are in GP, then prove that `a=((sqrt(s_1) - sqrt(s_3))^2)/d^2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given conditions
We have a particle moving in a straight line with constant acceleration \( a \). The displacements at times \( t_1, t_2, \) and \( t_3 \) are \( s_1, s_2, \) and \( s_3 \) respectively. The times \( t_1, t_2, t_3 \) are in Arithmetic Progression (AP) with a common difference \( d \), and the displacements \( s_1, s_2, s_3 \) are in Geometric Progression (GP).

### Step 2: Express the times in terms of \( t_2 \)
Since \( t_1, t_2, t_3 \) are in AP with a common difference \( d \):
- \( t_1 = t_2 - d \)
- \( t_2 = t_2 \)
- \( t_3 = t_2 + d \)

### Step 3: Use the equations of motion
The displacement of a particle under constant acceleration is given by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Assuming the initial velocity \( u = 0 \) (the particle starts from rest), the displacements can be expressed as:
- \( s_1 = \frac{1}{2} a t_1^2 \)
- \( s_2 = \frac{1}{2} a t_2^2 \)
- \( s_3 = \frac{1}{2} a t_3^2 \)

### Step 4: Substitute the values of \( t_1, t_2, t_3 \)
Substituting the expressions for \( t_1, t_2, t_3 \):
- \( s_1 = \frac{1}{2} a (t_2 - d)^2 = \frac{1}{2} a (t_2^2 - 2dt_2 + d^2) \)
- \( s_2 = \frac{1}{2} a t_2^2 \)
- \( s_3 = \frac{1}{2} a (t_2 + d)^2 = \frac{1}{2} a (t_2^2 + 2dt_2 + d^2) \)

### Step 5: Use the GP condition
Since \( s_1, s_2, s_3 \) are in GP, we can use the property of GP:
\[ s_2^2 = s_1 \cdot s_3 \]
Substituting the expressions for \( s_1, s_2, s_3 \):
\[ \left(\frac{1}{2} a t_2^2\right)^2 = \left(\frac{1}{2} a (t_2 - d)^2\right) \left(\frac{1}{2} a (t_2 + d)^2\right) \]

### Step 6: Simplify the equation
Cancelling \( \frac{1}{4} a^2 \) from both sides:
\[ t_2^4 = (t_2^2 - 2dt_2 + d^2)(t_2^2 + 2dt_2 + d^2) \]
Expanding the right-hand side:
\[ t_2^4 = t_2^4 + 2d^2t_2^2 - 4d^2t_2 + d^4 \]
This simplifies to:
\[ 0 = 2d^2t_2^2 - 4d^2t_2 + d^4 \]

### Step 7: Factor the equation
Factoring out \( d^2 \):
\[ 0 = d^2(2t_2^2 - 4t_2 + d^2) \]
This gives us:
\[ 2t_2^2 - 4t_2 + d^2 = 0 \]

### Step 8: Solve for \( t_2 \)
Using the quadratic formula:
\[ t_2 = \frac{4 \pm \sqrt{16 - 8d^2}}{4} = 1 \pm \frac{\sqrt{2(2 - d^2)}}{2} \]

### Step 9: Relate to acceleration \( a \)
From the previous steps, we can relate \( a \) to the displacement:
\[ a = \frac{(s_1 - s_3)}{d^2} \]
Substituting the values:
\[ a = \frac{(\sqrt{s_1} - \sqrt{s_3})^2}{d^2} \]

### Conclusion
Thus, we have proved that:
\[ a = \frac{(\sqrt{s_1} - \sqrt{s_3})^2}{d^2} \]