MHB Solve the nth term for the following list: 1/2, 5/6, 1, 1 1/10

[hm2018April]

Dear All

Please could you kindly help? I am new to this forum, kindly excuse me for asking.

The nth term for the following set of numbers is required. If someone could please guide me, I would be grateful.

Thank you.
Suggestion to solve the nth term suggestion for the following list please:
1/2, 5/6, 1, 11/10, ..., ...

 

[Greg]

Hello hm2018April and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[hm2018April]

Hello hm2018April and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Dear Contributor,

Thank you for the message.

Looking at the list of 1/2, 5/6, 1, 11/10, I converted them each respectively to the following fractions: 15/30, 25/30, 30/30, 31/30

I then focused on the numerator for each fraction.

I was able to work out the 5th term, which I believe is 35/30.

After that, working out the nth term is an area that I would like help with please.

Thank you.

 

[topsquark]

Dear Contributor,

Thank you for the message.

Looking at the list of 1/2, 5/6, 1, 11/10, I converted them each respectively to the following fractions: 15/30, 25/30, 30/30, 31/30

I then focused on the numerator for each fraction.

I was able to work out the 5th term, which I believe is 35/30.

After that, working out the nth term is an area that I would like help with please.

Thank you.

Can you give us any idea where the series comes from? Technically the next term in the series could be argued to be anything.

-Dan

 

[Opalg]

Suggestion to solve the nth term suggestion for the following list please:
1/2, 5/6, 1, 11/10, ..., ...

Hi hm2018April, and welcome to MHB!

Mathematicians hate problems like this, because there usually isn't a single, "correct" solution. You need to spot a pattern in the given sequence, and hope that it is the one envisaged by whoever set the problem.

What occurs to me about the sequence $$\frac12, \quad \frac56, \quad 1,\quad \frac{11}{10},\ \ldots$$ is that if you re-write it in the form $$\frac24, \quad \frac56, \quad \frac88,\quad \frac{11}{10},\ \ldots,$$ then the numerators and denominators are both increasing in a very predictable way. Can you use that to see how the sequence might continue, and provide a formula for the $n$th term?

 

[I like Serena]

Hi hm2018April, and welcome to MHB!

Mathematicians hate problems like this, because there usually isn't a single, "correct" solution. You need to spot a pattern in the given sequence, and hope that it is the one envisaged by whoever set the problem.

What occurs to me about the sequence $$\frac12, \quad \frac56, \quad 1,\quad \frac{11}{10},\ \ldots$$ is that if you re-write it in the form $$\frac24, \quad \frac56, \quad \frac88,\quad \frac{11}{10},\ \ldots,$$ then the numerators and denominators are both increasing in a very predictable way. Can you use that to see how the sequence might continue, and provide a formula for the $n$th term?

My own best guess was to look at the differences: $\frac 13, \frac 16, \frac 1{10}$.
The next difference might be $\frac 1{15}$, although it does seem a bit farfetched.
Your sequence looks much cleaner and more predictable.
Interestingly, the result for the next term is the same.

 

[hm2018April]

Can you give us any idea where the series comes from? Technically the next term in the series could be argued to be anything.

-Dan

Thank you so much for the reply. It was sourced from a Common Entrance Paper (13+). Thank you for asking.

 

[hm2018April]

Hi hm2018April, and welcome to MHB!

Thank you so much for the reply. It was wonderfully helpful.

Thank you.

 

[Opalg]

My own best guess was to look at the differences: $\frac 13, \frac 16, \frac 1{10}$.
The next difference might be $\frac 1{15}$, although it does seem a bit farfetched.
Your sequence looks much cleaner and more predictable.
Interestingly, the result for the next term is the same.

Not just for the next term, but for the whole sequence! In fact, your prediction and mine both say that the $(n+1)$th term minus the $n$th term is $\frac2{(n+1)(n+2)}$. So we are in agreement. (Handshake)

 

[hm2018April]

Not just for the next term, but for the whole sequence! In fact, your prediction and mine both say that the $(n+1)$th term minus the $n$th term is $\frac2{(n+1)(n+2)}$. So we are in agreement. (Handshake)

Thank you so much also. With best wishes.

 

[hm2018April]

Dear Contributors, I cannot express my thanks for all of your wonderful advice. Sorry for any delay. Thank you. I believe the answer is: 3n-1/2n+2

 

[topsquark]

Dear Contributors, I cannot express my thanks for all of your wonderful advice. Sorry for any delay. Thank you. I believe the answer is: 3n-1/2n+2

Tsk tsk. It would be (3n - 1)/(2n + 2). Parentheses are important.

-Dan

 

[hm2018April]

Tsk tsk. It would be (3n - 1)/(2n + 2). Parentheses are important.

-Dan

Dear Dan, Thank you for that. Sorry for the omission. Thank you.