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Simple question around Sequences and Series

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Simple question around Sequences and Series

by Balrog1978 » Fri Dec 24, 2010 8:25 pm
When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?

Answers: 3/10, 2/5, 1/2, 2/3, 6/5
To solve such problems, I've been used to the formula:
an = a0 + (n-1)d

Using this; I applied the two situations as:
a(4) = 4 + 3d
a(6) = 4 + 5d

thus : a(6) = a(4) + 1/5(a4)
solving these two I ended up getting d = 1/2.
the ANSWER is IN FACT d = 2/3!!!!

Turns out that the formula applies is IN FACT
a(n) = a0 + nd!!!

instead of
a(n) = a0 + (n-1)d!!!

Doesn't one take the 1st year as the ZEROth year? This caught me off guard. I'm missing something elementary here - Please someone clarify this for me!!! When to use nd vs (n-1)d?
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by Rahul@gurome » Fri Dec 24, 2010 8:43 pm
Solution:
Let the growth after every one year be g.
So height of tree h1 after 1 year is h1 = 4+g.
h2 = 4+2g.
So h4 = 4 + 4g and h6 = 4 + 6g.
Now h6 = (1 + 1/5)*h4 = 6/5*h4.
Or (4 + 6g) = (6/5)*(4+4g).
Or g = 2/3.

You have to take h1 = 4+g.

So h(n) = h1 + (n-1)*g = 4 + g + (n-1)*g = 4 + n*g.
Last edited by Rahul@gurome on Fri Dec 24, 2010 8:47 pm, edited 1 time in total.
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by anshumishra » Fri Dec 24, 2010 8:44 pm
Balrog1978 wrote:
When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?

Answers: 3/10, 2/5, 1/2, 2/3, 6/5
To solve such problems, I've been used to the formula:
an = a0 + (n-1)d

Using this; I applied the two situations as:
a(4) = 4 + 3d
a(6) = 4 + 5d

thus : a(6) = a(4) + 1/5(a4)
solving these two I ended up getting d = 1/2.
the ANSWER is IN FACT d = 2/3!!!!

Turns out that the formula applies is IN FACT
a(n) = a0 + nd!!!

instead of
a(n) = a0 + (n-1)d!!!

Doesn't one take the 1st year as the ZEROth year? This caught me off guard. I'm missing something elementary here - Please someone clarify this for me!!! When to use nd vs (n-1)d?
A(n) = A1 + (n-1) d -----> if you consider the first term to be A1

If you consider something like A(0) to be the first term,
Then A(1) = A(0) + d
A(2) = A(0) + 2d
......
A(n) = A(0) + nd [Please note that this n+1th term in this case]

Basically, all the series has common difference "d", Both the formulae, represent the same thing.
Thanks
Anshu

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by Balrog1978 » Fri Dec 24, 2010 9:39 pm
anshumishra wrote:
Balrog1978 wrote:
When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?

Answers: 3/10, 2/5, 1/2, 2/3, 6/5
To solve such problems, I've been used to the formula:
an = a0 + (n-1)d

Using this; I applied the two situations as:
a(4) = 4 + 3d
a(6) = 4 + 5d

thus : a(6) = a(4) + 1/5(a4)
solving these two I ended up getting d = 1/2.
the ANSWER is IN FACT d = 2/3!!!!

Turns out that the formula applies is IN FACT
a(n) = a0 + nd!!!

instead of
a(n) = a0 + (n-1)d!!!

Doesn't one take the 1st year as the ZEROth year? This caught me off guard. I'm missing something elementary here - Please someone clarify this for me!!! When to use nd vs (n-1)d?
A(n) = A1 + (n-1) d -----> if you consider the first term to be A1

If you consider something like A(0) to be the first term,
Then A(1) = A(0) + d
A(2) = A(0) + 2d
......
A(n) = A(0) + nd [Please note that this n+1th term in this case]

Basically, all the series has common difference "d", Both the formulae, represent the same thing.
Well - the thing that gets to me (been working hard at this one - so please bear with me). During the test - I would traditionally come across one of the two types of problems:

If we have a sequence like : 4, 6, 8, 10... and we're asked to calculate, say the 6th term.
then I would use a(n) = a0 + (n-1) d
= 4 + 5d
= 4 + (5)2 = 14 - which is correct

However, in problems such as the one I've mentioned in this thread, I did a (natural to me - absurd to others I'm pretty sure).
a6 = a0 + (n-1)d
= 4 + 5d
which gave me 1/2 as I described in my original posting!!!!

Its this "natural but incorrect" switch between the two types that caused me (and continues to cause me) these kinds of mistakes, especially under test pressure. So if the two are indeed the same, how do I reinterpret the question in THIS thread to use the (n-1) logic?
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by anshumishra » Fri Dec 24, 2010 9:55 pm
Balrog1978 wrote:
anshumishra wrote:
Balrog1978 wrote:
When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?

Answers: 3/10, 2/5, 1/2, 2/3, 6/5
To solve such problems, I've been used to the formula:
an = a0 + (n-1)d

Using this; I applied the two situations as:
a(4) = 4 + 3d
a(6) = 4 + 5d

thus : a(6) = a(4) + 1/5(a4)
solving these two I ended up getting d = 1/2.
the ANSWER is IN FACT d = 2/3!!!!

Turns out that the formula applies is IN FACT
a(n) = a0 + nd!!!

instead of
a(n) = a0 + (n-1)d!!!

Doesn't one take the 1st year as the ZEROth year? This caught me off guard. I'm missing something elementary here - Please someone clarify this for me!!! When to use nd vs (n-1)d?
A(n) = A1 + (n-1) d -----> if you consider the first term to be A1

If you consider something like A(0) to be the first term,
Then A(1) = A(0) + d
A(2) = A(0) + 2d
......
A(n) = A(0) + nd [Please note that this n+1th term in this case]

Basically, all the series has common difference "d", Both the formulae, represent the same thing.
Well - the thing that gets to me (been working hard at this one - so please bear with me). During the test - I would traditionally come across one of the two types of problems:

If we have a sequence like : 4, 6, 8, 10... and we're asked to calculate, say the 6th term.
then I would use a(n) = a0 + (n-1) d
= 4 + 5d
= 4 + (5)2 = 14 - which is correct

However, in problems such as the one I've mentioned in this thread, I did a (natural to me - absurd to others I'm pretty sure).
a6 = a0 + (n-1)d
= 4 + 5d
which gave me 1/2 as I described in my original posting!!!!

Its this "natural but incorrect" switch between the two types that caused me (and continues to cause me) these kinds of mistakes, especially under test pressure. So if the two are indeed the same, how do I reinterpret the question in THIS thread to use the (n-1) logic?
I understood your problem now. For the current problem, what should be the height of the tree after first year (at the end of first year which one can also confuse with at the beginning of 2nd year..This may be your problem):
a+d
After 2nd year : a+2d...
After 4th year : a+4d

You should be careful to lookout for these cases OR (if it is so impossible) just don't use the a(n) representation (Use X and Y). You can still represent the equation in terms of X and Y in terms of d. See if it helps !
Thanks
Anshu

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by Balrog1978 » Sat Dec 25, 2010 8:43 am
I see what you're saying. "Don't go in guns blazing!" :)
So effectively - the PRIMER - how to identify the first couple of sequences should be logically inferred.
A blind jump to the formula should never be the first step!!

To sum it up though :
1) "hard sequences" - such as the one I mentioned : 4,6,8.... the n-1 tactic can straightaway be used
2) "Inferred sequences" - such as word problems in this thread - ARRIVE at the first couple of sequences and then BUILD the general 'n' or 'n-1' term in the series!!

This about right?
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by anshumishra » Sat Dec 25, 2010 8:46 am
Balrog1978 wrote:I see what you're saying. "Don't go in guns blazing!" :)
So effectively - the PRIMER - how to identify the first couple of sequences should be logically inferred.
A blind jump to the formula should never be the first step!!

To sum it up though :
1) "hard sequences" - such as the one I mentioned : 4,6,8.... the n-1 tactic can straightaway be used
2) "Inferred sequences" - such as word problems in this thread - ARRIVE at the first couple of sequences and then BUILD the general 'n' or 'n-1' term in the series!!

This about right?
Yes, you are right !
From your post I was able to deduce that you know the concept well, so be a little cautious to make sure that the formula is what it should be.
Thanks
Anshu

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by Balrog1978 » Sat Dec 25, 2010 10:03 am
Thanks!!! This was a lot of help!!!
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