[AS Level] - Geometric Progressions.

[Guest]

I've been struggling with geometric progressions for a while now, but I still can't quite get my head around how to apply the equations and logic to real situations (which are most commonly applied in our tests to sequences/progressions). Does anyone have any tips for dealing with them, or likewise?

An example of a question I've been finding difficult:

A competitor is running in a 25 km race. For the first 15 km, she runs at a steady rate of 12 kmh-1. After completing 15 km, she slows down and it is now observed that she takes 20% longer to complete each kilometre than she took to complete the previous kilometre.*

a) Find the time, in hours and minutes, the competitor takes to complete the first 16 km of the race.*

This I managed to get, through (15/12) + 1/(12-0.2(12)) = 1.35416... = 1 hour, 21 minutes.

The time taken to complete the rth kilometre is Ur hours.

b) Show that, for 16 =< r =< 25, Ur = 1/12*(1.2)r - 15

c) Using the answer to b, or otherwise, find the time, to the nearest minute, that she takes to complete the race.

There are quite a few other instances of people asking for help with this, but I don't really understand the method. I would really appreciate it if anybody has any advice for appropriately converting situations with distances or money into a better format - thanks!

[HallsofIvy]

A competitor is running in a 25 km race. For the first 15 km, she runs at a steady rate of 12 kmh-1. After completing 15 km, she slows down and it is now observed that she takes 20% longer to complete each kilometre than she took to complete the previous kilometre.{/quote]
15 km at 12 km/h takes 15/12= 5/4 hour= 75 minutes. Also, at 12 km/h one km takes 1/12 hour= 60/12 minutes= 5 minutes.

a) Find the time, in hours and minutes, the competitor takes to complete the first 16 km of the race.*

This I managed to get, through (15/12) + 1/(12-0.2(12)) = 1.35416... = 1 hour, 21 minutes.

Yes. She takes 20% longer to finish the 16 hour than the 15th hour so 1.2(5)= 6 minutes to finish the 16th km. She takes 75+ 6= 81 minutes which is 60+ 21= one hour and 21 minutes.

The time taken to complete the rth kilometre is Ur hours.

b) Show that, for 16 =< r =< 25, Ur = 1/12*(1.2)r - 15

As above, it takes 6 minutes to run the 16 km, 6(1.2)= 7.2 minutes to run the 17 km, 7.2(1.2)= 8.64 minutes to run the 18th km. For each new kilometer, we multiply by 1.2 again.
The formula you give, if read literally, would give (1/12)(1.2)(18)- 15= 1.8- 16= -14.2 minutes! You need parentheses around (r- 15)! But even that would give (1/12)(1.2)(18- 15)= 3/10= 0.3 minutes. IF r> 15 then it takes $6(1.2)^{r- 15}$ minutes to run the rth mile. You should clearly state that this formula works only for r> 15. The time to run each kilometer for r< 16 is 6 minutes.

c) Using the answer to b, or otherwise, find the time, to the nearest minute, that she takes to complete the race.

To run the entire 25 kilometers requires the initial one hour and 21 minutes plus 6 minutes plus 7.2 minutes plus 8.64 minutes plus ... plus $6(1.2)^(r- 15)$ for r from 16 to 25. Letting i= r- 15 that last sum (not including the initial one hour and 21 minutes) is $6(1.2)+ 6(1.2)^2+ ...+ 6(1.2)^10= 6(1.2)[1+ 1.2+ (1.2)^2+ ...+ (1.2)^9]$. That is 6(1.2)= 7.2 times the geometric sum $1+ 1.2+ (1.2)^2+ ...+ (1.2)^9$. Don't forget to add the initial one hour and 21 one minutes for the first 15 km.