# progression

• Jun 11th 2008, 08:17 AM
bongz
progression
a tree is planted when it has a height of 2 metres. duringthe first year it grows to a height of 3 metres. thereafter the increase in height every year is 4/5 of the increase in height durind the previus year. what isthe maximum height that the tree will reach?(Thinking) please people help me out on this one, its a sequence and series question
• Jun 11th 2008, 08:33 AM
TKHunny
So, write it out and ponder it...

$2 + 1 + 4/5 + (4/5)^{2} + (4/5)^{3} + ... = 2 + \left(\frac{1}{1-\frac{4}{5}}\right)$

You may be expected to 1) Recognize the structure of the Geometric Series and 2) Know the formula for the sum.

Note: The question is a little funny. I don't really like the phrase "maximum height that the tree will reach". I would prefer a limit sort of phrase, maybe something like "What height will the tree approach?" or "What is the minimum height the tree will NOT exceed?" It is a tough concept to get exactly right. Maybe the original question is okay for now.
• Jun 11th 2008, 08:48 AM
bongz
thanx 4 helping(Clapping), however. i am still puzzled on how u got the answer(Worried)
• Jun 11th 2008, 10:07 AM
Moo
Hello,

Allow me to answer...

Quote:

Originally Posted by TKHunny
So, write it out and ponder it...

$2 + 1 + 4/5 + (4/5)^{2} + (4/5)^{3} + ... = 2 + \left(\frac{1}{1-\frac{4}{5}}\right)$

You may be expected to 1) Recognize the structure of the Geometric Series and 2) Know the formula for the sum.

Note: The question is a little funny. I don't really like the phrase "maximum height that the tree will reach". I would prefer a limit sort of phrase, maybe something like "What height will the tree approach?" or "What is the minimum height the tree will NOT exceed?" It is a tough concept to get exactly right. Maybe the original question is okay for now.

Quote:

Originally Posted by bongz
thanx 4 helping(Clapping), however. i am still puzzled on how u got the answer(Worried)

Do you understand the first part ?

--> $h=2+1+(4/5)+(4/5)^2+\dots$

You can see that 2 is the original height. The following part is a geometric series.

$1+q+q^2+q^3+\dots+q^n=\frac{1-q^{n+1}}{1-q}$

Here, $n \to \infty$
And $|q|<1$, which means that $\lim_{n \to \infty} q^n=0$

So $1+q+q^2+q^3+\dots \to \infty=\frac{1}{1-q}$

Is it more clear ? :x
• Jun 11th 2008, 02:31 PM
TKHunny
Quote:

Originally Posted by bongz
thanx 4 helping(Clapping), however. i am still puzzled on how u got the answer(Worried)

1) I didn't get the answer. I left that for you.

2) I gave you a list of two things you are expected to know. Look them up and start knowing them. Only then will you see how YOU will get such answers.