Inspired my myself!

Let's look at the $x$ the $y$ separately.:

The initial $x$ value is 3, and because the particle only moves horizontally every other iteration and alternating positive and negative direction, the change in $x$ can be represented by the geometric series $\sum_{n=0}^\infty 10(\frac {-1}{4})^{n}$

This series converges to $\frac {10}{1-\frac {-1}{4}}$ which simplifies to $8$ , so our final $x$ coordinate is $3 + 8 = 11$

The initial $y$ value is 5, and for the same reasons as above, the change in $y$ can be represented by $\sum_{n=0}^\infty -5(\frac {-1}{4})^n$

This series converges to $\frac {-5}{1-\frac {-1}{4}}$ which simplifies to $-4$ , so our final $y$ coordinate is $5 - 4 = 1$

Therefore, the particle ends up at $(11,1)$

To find the distance from the origin to this point we just use the distance formula.

$\sqrt{(11-0)^{2} + (1 - 0)^{2}} = \sqrt{122}$ , so our answer is $\boxed{122}$

Start the process with respect to origin(i.e let the particle started from origin instead of (3,5)).Now work for the final X- Coordinate and final Y-Coordinate of the particle independently.Final X coordinate is the sum of the infinite geometric progression 10,-10/4,10/16,-10/64........=10/(1+1/4)=8 .Similarly, Final Y-Coordinate is the sum of the infinite GP 5,-5/4,5/16.........=4. But the particle started towards right with initial X-Coordinate 3 => actual final X-Coordinate=3+8=11. And actual final Y-Coordinate =5-4 =1 as the particle started down with respect to (3,5) initially.Hence required 'x'=11^2+1^2=122

Solving graphically by simultaneous equations and distance formula

That was really a nice problem...

Let the coordinates of the final point be $P(X,Y)$ . Therefore, the distance of the particle at the final point P from the origin O is $OP=\sqrt{X^2+Y^2}=\sqrt{x} \Rightarrow x=X^2+Y^2$ .

Therefore, we are supposed to enter the value of $X^2+Y^2$ as the answer.

$(3,5)$ are the coordinates of the initial point.

The Coordinates change as the particle moves, the x-coordinate varies as $X = 3+10-\frac{10}{2^2}+\frac{10}{2^4}-\frac{10}{2^6}+\ldots \infty$ $X = 3+\text{Infinite Converging G.P. Series with the First term, a=10, Common Ratio, r=-1/4}$ $X = 3+a\frac{1}{1-r}=3+10\frac{1}{1-(-\frac{1}{4})}=3+8$ $\Rightarrow X = 11$

Similarly, $Y=y_0-\frac{10}{2}+\frac{10}{2^3}-\frac{10}{2^5}+\ldots \infty$ $Y=5-\frac{10}{2} \times \frac{1}{1-\frac{1}{4}}=5-4$ $\Rightarrow Y = 1$

Therefore, $x=X^2+Y^2=11^2+1^2=\boxed{122}$ :D

P.S. Change in Coordinates of the position of the particle explained : As the particle moves firstly 10 units along the positive x direction, x coordinate increases by 10 units. Further when the particle turns to the right i.e. towards negative y direction and moves $10/2=5$ units, the y co-ordinate decreases by 5 units. Now as the particle turns to the right, it faces the negative x direction, and moves by $10/(2^2)=2.5$ units, the x-coordinate decreases by 2.5 units. The process goes on. The changes can be listed as (x co-ordinate increases by 10 units, y co-ordinate decreases by 5 units, x co-ordinate decreases by 2.5 units,...) Clubbing changes in x coordinate together, we get (x co-ordinate increases by 10 units, x co-ordinate decreases by 2.5 units, ...). Clubbing changes in y coordinate together, we get (y co-ordinate decreases by 5 units, y co-ordinate increases by 1.25 units, ...). Applying these changes to the initial coordinates, gives us the position of the final point.