The value of $(a+b)$, according to the question. 
My friend gave me a question I tried my best, but I'm low on triangle concept.
Points $ O, A, B, C... $ are shown in the figure where $ OA=2AB=4BC=...$ and so on. Let $A$ be the centroid of a triangle whose orthocentre and circumcentre are $(2,4)$ and $(\frac72,\frac52)$, respectively, if an insect starts moving from the point $O=(0,0)$ along the straight line in zig zag fashions and terminates ultimately at point $P(a,b)$, then find the value of $(a+b)$.
I tried using the collinearity property of centroid, circumcentre and orhtocentre, and the distance property also, but reached nowhere. Please help.
 A: From basic trigonometry we get that that the $x$ coordinates tends to...
$$a=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\cos (45)$$
Here $\frac{d}{2^n}$ denotes the magnitude of the $n+1$th hypotenuse (we start the sum at $0$ that is why it's a bit weird with $n+1$). How did I get this? You implied the distances (hypotenuses) went $d,d/2,d/4$ and I assumed this was a geometric progression.  Where here $d$ is magnitude of the first hypotenuse $n=0$, aka the distance between the origin and point A. Using the assumption you will get the formula for the hypotenuse in terms of $n$. Which you can then use to find the horizontal components of each component then add them up to get $a$ as they are all positive.
Using the same method (but with vertical components alternating direction/sign), the $y$ coordinate tends to:
$$b=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\sin (45) (-1)^{n}$$
Which converges to a positive number. 
Note $\cos (45) = \sin (45) = \frac{1}{\sqrt{2}}$
For the first sum,
$$\sum_{n=0}^{\infty} (\frac{1}{2})^n=\frac{1}{1-(1/2)}=2$$
So as a result $a=\frac{2d}{\sqrt{2}}=\sqrt{2} d$.
For the second one, 
$$\sum_{n=0}^{\infty}\frac{ (-1)^{n}}{2^n}=\sum_{n=0}^{\infty} (\frac{-1}{2})^n=\frac{1}{1-(-1/2)}=\frac{2}{3}$$
From there,
$$b=\lim_{N \to \infty} \sum_{n=0}^{N} \frac{d}{2^n}\sin (45) (-1)^{n}=\frac{2d}{3 \sqrt{2}}=\frac{\sqrt{2}d}{3}$$
$$a+b=\frac{4d\sqrt{2}}{3}$$
Now all that is left is to find $d$....
Finding $d$ has to do with the Euler line @almagest, which passes through the orthocenter, centroid, and circumcenter. This and the fact that $A$ forms a 45-45-90 isosceles triangle with the $x$ axis so it's coordinates are $(x_0,y_0)=(x_0,x_0)$. Let's find a formula that describes the Euler line given our two points $(2,4)$ and $(3.5,2.5)$.
The slope of the line is:
$$\frac{4-2.5}{2-3.5}=-1$$
So the equation of the line is:
$y=-1(x-2)+4=-x+6$
Substituting the point $A=(x_0,x_0)$  in we get:
$x_0=-x_0+6$
$x_0=3$
Thus,
$$d=\sqrt{x_0^2+x_0^2}=\sqrt{2}|x_0|=3\sqrt{2}$$
And thus,
$$a+b=\frac{4(3\sqrt{2})\sqrt{2}}{3}=8$$
Also
Plugging $d$ back into the formulas we got for $a$ and $b$ you may get that $P=(6,2)$.
A: A less complicated method that builds off of @Ahmed S. Attaalla's solution. 
Because of the Euler line, we know that the centroid is 1/3 of the way between the circumcenter and the orthocenter. 

In other words, $A=\displaystyle\frac{2*\left (\frac{7}{2},\frac{5}{2}\right)+(2,4)}{3}=\left(3,3\right)$, because $A$ is $2$ times closer to the circumcenter than the orthocenter (and on the same line).
Now, the $x$-coordinate of point P is $\displaystyle \sum_{n=0}^{\infty}\frac{3}{2^n}=6$, and the $y$-coordinate is $\displaystyle \sum_{n=0}^{\infty}\frac{3}{(-2)^n}=2.$
So the final answer is $6+2=8$.
A: If the two vectors are $v_1$ and $v_2$, your description is something like:
$$v_{tot} = v_1 + \frac12 v_2 + \frac14 v_1 + \frac18 v_2 \dots$$
This is equivalent to:
$$\begin{align}
v_{tot} & = (v_1 + \frac12 v_2) + \frac14(v_1 + \frac12 v_2) \dots\\
&= (v_1 + \frac12 v_2)(1 + \frac14 + \frac1{16}\dots)\\
\end{align}$$
The geometric series is equal to $\frac{1}{1 - \frac14} = \frac43$. So once you figure out $v_1$ and $v_2$, your final point will be
$$\frac43(v_1 + \frac12v_2)$$
