Does this question involve series? While solving the problem 


Half the people on a bus get off at each stop after the first and no one gets on after the first stop.If only one person gets off at stop number 7 .How many people got on at the first stop ?


Here is how I am solving it but I guess I am wrong
Let no of people who got on = x
so $x + \frac{x}{2} + \frac{x}{4}+ \frac{x}{8}+ \frac{x}{16}+ \frac{x}{32}+ 1 = 0 $ 
Any suggestions ?
 A: Let us do it in your style.  Say that there are $x$ people on the bus after the first stop.  
At the second stop, $\frac{x}{2}$ get off, so $\frac{x}{2}$ are left.
At the third stop, $\frac{x}{4}$ get off, and $\frac{x}{4}$ are left.
At the fourth stop, $\frac{x}{8}$ get off, $\frac{x}{8}$ are left.
At the fifth stop, $\frac{x}{16}$ get off, $\frac{x}{16}$ are left.
At the sixth stop, $\frac{x}{32}$ get off, $\frac{x}{32}$ are left.
At the seventh stop, $\frac{x}{64}$ get off, and $\frac{x}{64}$ are left.
But $1$ person is left. So $\frac{x}{64}=1$, and therefore $x=64$.
Alternately, there were $2$ people on the bus just before the $7$th stop, so $4$ just before the sixth, so $8$ just before the fifth, so $16$just before the fourth, so $32$ just before the third, so $64$ just before the second stop.
Or else, (but this is too much work, $\frac{x}{2}$ got off at the second, and so on, leaving $1$ person on the bus after the seventh.  This gives us the equation
$$x=\frac{x}{2}+\frac{x}{4}+\frac{x}{8}+\frac{x}{16}+\frac{x}{32}+\frac{x}{64}+1.$$
Solve for $x$. Maybe to make life easier first multiply both sides by $64$.
Remark: Does this question involve series? Well, if we solve it my third way, which is fairly inefficient, then indeed we end up finding the sum of a finite geometric series. But there are more efficient ways that just involve a sequence.
