Help with sequences?

**Fiasco** · December 4th, 2011, 06:42 AM

Hello, I'm in need of some help for my math homework as follows:

Consider the two sequences 4, 8, 14, 22, 32... and 5000, 5200, 5400, 5600, 5800, 6000... How many terms will it take for the first sequence to overtake the second sequence?

I got the rule for both sequences but they didn't seem to help me solve it.

Thanks in advance.

**skeeter** · December 4th, 2011, 06:50 AM

Originally Posted by Fiasco

Hello, I'm in need of some help for my math homework as follows:

Consider the two sequences 4, 8, 14, 22, 32... and 5000, 5200, 5400, 5600, 5800, 6000... How many terms will it take for the first sequence to overtake the second sequence?

I got the rule for both sequences but they didn't seem to help me solve it.

Thanks in advance.

use the rule you determined for the first sequence and set it equal to 5000

**Fiasco** · December 4th, 2011, 06:58 AM

So, if the rule for the first sequence is:

y = 7/5 x + 12/5

I make it 5000 = 7/5 x + 12/5 ?

EDIT:

The rule I found for the second sequence is:

y = x + 200

**skeeter** · December 4th, 2011, 07:09 AM

Originally Posted by Fiasco

So, if the rule for the first sequence is:

y = 7/5 x + 12/5

I make it 5000 = 7/5 x + 12/5 ?

EDIT:

The rule I found for the second sequence is:

y = x + 200

the rle for the first sequence is $a_n = n^2 + n + 2$

the rule for the second sequence is $b_n = 5000 + 200(n-1)$

**Fiasco** · December 4th, 2011, 07:15 AM

Well, I'm lost. Anyways, thanks for the help. I guess my definition of rule is really different from yours. I'll just ask my classmates help me.

**e^(i*pi)** · December 4th, 2011, 07:16 AM

Originally Posted by Fiasco

So, if the rule for the first sequence is:

y = 7/5 x + 12/5

I make it 5000 = 7/5 x + 12/5 ?

Let's pick a value of x and see. To make life easy we'll take the fifth term (x=5) which we know to be 32 from the question: $\frac{7}{5} \cdot 5 + \dfrac{12}{5} = 7 + \frac{12}{5} \neq 32$ so that's not the sequence.

To find the formula for the first equation look at the differences between terms. The second difference is the change between the first differences/changes.

First Differences: 4, 6, 8, 10
Second Differences: 2, 2, 2, 2.

The fact that the second differences are all the same value means we have an equation of order two which is quadratic. By trial and error I get $u_{n} = n^2+n+2$ for the nth term of your first sequence (perhaps someone knows a better way?)

EDIT:

The rule I found for the second sequence is:

y = x + 200

The second sequence is an arithmetic sequence with first term 5000 and common difference 200 so it's nth term is given by $u_n = 5000+200(n-1)$

I suspect the question is saying that the second term is increasing along with the first term for values of n so we need to take that into account (meaning you can't set it to 5000 and be done with it). Instead set the two equations equal to each other $n^2+n+2 = 5000+200(n-1)$ and solve for n remembering that $n > 0$

The first sequence will grow quite slowly at first so don't be surprised if you get a high value

**HallsofIvy** · December 4th, 2011, 07:18 AM

fiasco, your formulas make no sense at all. What does "x" represent?

**skeeter** · December 4th, 2011, 07:34 AM

learn something about sequences ...

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