Geometric Progression Word Question I have been trying to complete this question for a while now, and can't seem to figure it out. Could any one please give me some assistance? Thanks heaps.
Arithmetic Progression Formulas
$$t_n=t_1+(n−1)$$ 
$$S_n=\frac{n}{2}[2t_1 +(n-1)d]$$
Geomtric Progession Formulas
$$t_n = t_1 ⋅r^{n-1}$$
$$S_n = \frac {t_1(r^n-1)}{r-1}$$
Question:⋅
Callum has to walk from the house to the far paddock. He takes his dog with him, but when he is 36 meters from the house the dog runs from Callum to the house and back again. Each time the dog gets to Callum it does it again. The dog runs 4 times as fast as Callum is walking and does this 7 times before Callum reaches the far paddock. How far from the house is the paddock?
Answer: About 1286 meters.
 A: It's not entirely obvious which formula (geometric/arithmetic/etc) to use at the beginning of the problem, so it makes sense to do some scratch work to figure out what is going on.
The $7$ times that the dog runs back and fourth from the house make things complicated.  What happens in the first run to the house?
Callum and the dog are both $36$ meters from the house.  The dog runs to the house ($36$ meters) and then back to Callum.  Since Callum has continued to walk this is $36$ meters plus the extra distance that Callum has walked.
Let $d$ be the distance that Callum walks, then the distance the dog goes is $72+d$.  Since the dog goes at $4$ times Callum's speed, this means that $4d=72+d$.  Hence, $d=24$ meters.  Therefore, the second run starts at $36+24=60$ meters from the house.
Now, you could replicate this calculation $6$ more times or (better) try to generalize.  If Callum and the dog are at a distance of $h$ from the house, how far does Callum walk in one run of the dog?
A: Following what Michael Burr said.
Therefore, 
the distance the dog now has to travel is $4d = 120 + d$. Thus, d = 40 meters.
We can determine the common ration using this formula (We now know that the question requires geometric progression not arithmetic progression.)
$$r = \frac{t_n}{t_(n-1)}$$
Hence $ \frac{60}{36} = \frac{100}{60} = r = 1 \frac {2}{3}$
Now, $t_7 = t_1 \times (r)^7$
$t_7 =36\times (1\frac {2}{3})^7$
Then $t_7 = 1286.00823m$ 
