Frog-Jumping Out Of Well Word Problem A frog is jumping out of a well 30 ft. deep. Each day he jumps 3 feet and slips two feet back.  How many days does it take the frog to jump up to 30 ft. (out of the well)?
I did this problem long-hand and got 28 days--an answer others got as well.  I am wondering if there is some series (Geometric series, etc.) I tried to use a "closed form" Geometric Series, but that does not seem to work. Any other series solutions other than long-hand.  Thanks.
 A: If you insist to solve the problem using a "series method", here's one way to do it.
We will use the Arithmetic progression to solve it.
Remember, Sum of AP to $n$ terms, $\bf S_n= \frac n2 [2a+(n-1)d]$ where $d$ is the common difference and $a$ is the first term.
The series are as below: 
$$S_n=3+3+3+3+3+\dots \text { (here } a=3 \text { and }d=0)$$
$$S_m=-2-2-2-2-2-\dots \text { (here } a=-2 \text { and }d=0)$$
We need to find the following: $$S_n+S_m=30$$
We have 2 cases:

*

*$n-m=1$ (When frog comes out before sliding down the slippery wall)

*$n=m$ (There's tunnel in well & frog's convinced to move out buttock first)

Case I:
Here, $n-m=1 \Rightarrow \bf m=n-1$
$$S_n+S_m=30$$
$$\Rightarrow \frac n2 [2(3)] + \frac {n-1}2 [2(-2)]=30 \text {  (as }d=0)$$
$$\Rightarrow 3n-2n+2=30$$
$$\Rightarrow \bf n=28$$
Case II:
Here, $\bf n=m$
$$S_n+S_m=30$$
$$\Rightarrow \frac n2 [2(3)] + \frac n2 [2(-2)]=30 \text {  (as }d=0)$$
$$\Rightarrow 3n-2n=30$$
$$\Rightarrow \bf n=30$$
But since we are interested only in Case I, so, it will take $28$ days for the frog to leap out of the well.
A: The frog progresses $3-2=1$ foot every day that it does not make it out. It makes it out once it jumps from $27$ feet. So it will take $28$ days.
Alternatively; the frog starts by jumping $3$ feet on the first day. From then on, the frog progresses by $-2+3=1$ foot every day. Hence it takes $28$ days to make it up to $30$ feet.
A: Alternatively, let $D(n)$ be the distance from the frogg to the top of the well on day $n^{th}$. We have: $D(1) = 30 - 3 + 2 = 29, D(2) = 29 - 3 + 2 = 28\implies D(n) = 30-n$, and $D(27) = 3$. So it needs one more day to get out of the well. So the number of days it needs to get out of the well is $27+1 = 28$.
A: If the well is 30 meters deep.. and the frog climbs 3meters day time and slips 2 meters night.this means it climbs 1m per day(24 hours). On the 27th day it wil only achieve 27m, so on the 28th day it will make a 3m climb which can only get the frog at the top edge of the well where it indicates 30meter.
According to the equation, since the frog only has the ability to make a 3 meter climb during the day, this mathematical keeps the frogs at the 30meter edge of the deep well, as futher movement in the sense of going out will be against the equation that states (3m in the day and 2m slip down in the Night).
This now Means that the frog would remain at the edge of the well till night thereby slipping down to 28meters at night and the next day being the 29th day.. it will make another 3m climb which would take it just 2 meters to get to the top edge of the well and the remaining 1meter mathematically permits the frog to make the get out of the well move.
So the answer is 29days.
