Reach 20m with 2 steps forward and 1 step back

[Guest]

My question is simple but I just need to double check it with you guys..,

A person wants to reach 20m. His first 2 steps reach 1 metre and then his next step goes back half a metre
that is to 1.5 metres, he then takes 2 steps that is 1 more metre from where he was and then goes back
half a metre...if he keeps doing this when will he reach 20 metres.

This is a sum to infinity question right?

Steps-----------------------Distance
2=============== 1
3=============== 1/2
5=============== 3/2
6=============== 4/2
8=============== 4
9=============== 7/2
11============== 9/2
13============== 9


So what is the formula I use?

 

[skeeter]

I'm afraid you don't have the correct idea ... you are confusing the concept of distance and displacement. Also, this is not a sum to infinity.

2 steps forward ... the person is at the position x = 1.0 m
1 step back ... the person is now at the position x = 0.5 m, not 1.5 m

2 steps forward ... now at x = 1.5 m
1 step back ... now at x = 1.0 m

get the idea?

 

[soroban]

Hello, americo74!

This is a classic trick question . . . usually with "a frog in a well."

A person wants to reach 20m.
His first 2 steps reach 1 metre and then his next step goes back half a metre; that is, to 1.5 metres.
He then takes 2 steps that is 1 more metre from where he was and then goes back half a metre.
If he keeps doing this, when will he reach 20 metres?

For two steps, he advances a half-meter each.
. . On his third step, he retreats a half-meter.
In other words, he advances a half-meter for every three steps.

To move 20 meters, he must advance 40 half-meters.
. . At a half-meter every 3 steps, it should take: \(\displaystyle 3\,\times\,40\:=\:120\) steps.
But this is wrong!

Why?
After 114 steps, he has advanced \(\displaystyle \frac{114}{3}\,=\,38\) half-meters \(\displaystyle = \,19\) meters.

In his next two steps, he advances another meter and reaches the 20-meter goal.

Therefore, it takes him 116 steps.