Question

An athlete us jumping. However, everytime she jumps she gets a bit more tired, and every jump goes 1/2 as far as her prior jump. Now, for her very first jump, she goes 1/2 of a foot. On her second jump, she goes 1/4 of a foot, and so on and so forth. How many jumps does it take to get to travel 1 foot

Answer 1

Answer:

She can never travel a distance of 1 foot for the given condition.

Step-by-step explanation:

In every jump, she goes 1/2 as far as her prior jump.

In the 1st jump, she covered a distance, d_1= \frac 1 2 foot.

The distance covered in the 2nd jump

$d_2= \frac 12 \times d_1 = = \frac 12 \times \frac 12= \frac 1 4$ (given)

So, the distanve covered in the 3rd jump

$=\frac 12 \times d_2= \left(\frac 1 2\right )^2\times \frac 1 2= \left(\frac 1 2\right)^{3-1}\times \frac 1 2.$

Similarly, the distance covered in the $r^{th}$ jump

$=\left(\frac 12 \right)^{r-1}\times \frac 1 2.$

Assuming she requires $n$ jumps to travel 1 foot of distance.

So, the sum of all the distances covered in n jumps = 1 foot

$\Rightarrow \frac 1 2 + \frac 1 4 + \cdots + \left(\frac 12 \right)^{r-1}\times \frac 1 2+\cdots+ \left(\frac 12 \right)^{n-1}\times \frac 1 2=1$

$\Rightarrow \frac 1 2 + \left(\frac 12 \right)^2 + \cdots + \left(\frac 12 \right)^{r}+\cdots+ \left(\frac 12 \right)^{n}=1$

This a geometric progression, G.P., of $n$ terms having common ration, r= 1/2 and the first term $a_1= 1/2$.

As sum of all the n terms of  G.P $=\frac {a_1(r^n-1)}{r-1}$,

$\Rightarrow \frac {1/2((1/2)^n-1)}{1/2-1}=1$

$\Rightarrow - \left(\left(\frac 1 2\right)^n-1\right)=1$

$\Rightarrow \left(\frac 1 2\right)^n-1=-1$

$\Rightarrow \left(\frac 1 2\right)^n=-1+1=0$

$\Rightarrow \frac {1}{2^n}=0$

This is only possible, mathematically, when $n\rightarrow \infty$, But in real life situation reaching infinity is not possible.

Hence, she can never travel a distance of 1 foot for the given condition.