In the following number sequence the interval between each pair of adjacent 
numbers decreases by one as the sequences progresses.

2    10    17    23    28

so...

 2 >>> 10 = +8
10 >>> 17 = +7
17 >>> 23 = +6
23 >>> 28 = +5

but ...

is there one constant rule that can be applied to any of the numbers to 
arrive at the next number in the sequence?

Thanks

Ahoyhoy

Ahoyhoy wrote:
> In the following number sequence the interval between each pair of
> adjacent numbers decreases by one as the sequences progresses.
>
> 2    10    17    23    28
>
> so...
>
> 2 >>> 10 = +8
> 10 >>> 17 = +7
> 17 >>> 23 = +6
> 23 >>> 28 = +5
>
> but ...
>
> is there one constant rule that can be applied to any of the numbers
> to arrive at the next number in the sequence?
>
> Thanks
>

Yes:
a + (n-1)xd + 0.5(n-1)(n-2)xC

Where a = the first term = 2

d= the first difference = 8

C = the difference increase = -1

So the general rule =  2 + (n-1)x8 + 0.5(n-1)(n-2)x -1

= 2+ (n-1)x8 - 0.5(n-1)(n-2)


sheri

"Ahoyhoy"  wrote in message 
news:goUJi.52991$rr5.42721@newsfe1-win.ntli.net...
> In the following number sequence the interval between each pair of 
> adjacent numbers decreases by one as the sequences progresses.
>
> 2    10    17    23    28
>
> so...
>
> 2 >>> 10 = +8
> 10 >>> 17 = +7
> 17 >>> 23 = +6
> 23 >>> 28 = +5
>
> but ...
>
> is there one constant rule that can be applied to any of the numbers to 
> arrive at the next number in the sequence?

You need to find the "general term", sometimes called the nth term and given 
the symbol T(n). n is the position in the sequence.

You've made a good start, using the differences is key finding the nth term.

If the first difference is a constant, the sequence is linear and T(n) = An 
+ B, with A and B constants to be found.

 If not, find the second difference ((ie the difference between the 1st 
differences)

If the second difference  is constant the sequence is quadratic
and T(n) = An^2+Bn+C,  with A, B, and C constants to be found.

A is second difference  / 2 so, in this case,  A = -0.5

You now know that  T(n) = -0.5n^2 + Bn + C

B and C can be found in several ways, I recommend rearranging  T(n) 
= -0.5n^2 + Bn + C
to give:

T(n) +0.5n^2=  Bn + C

This shows, if you take each term of the original sequence and add 0.5n^2 to 
it, you will generate a linear sequence  Bn+C

You can find B and C by the usual ways (difference for B and "zeroth term" 
for C)

You then have T(n) for your sequence and can check it using the terms you 
have. After that, you can find any term by just subing in n

Brian

On 25 Sep, 20:42, "Brian Reay" <s...@website.invalid> wrote:
> "Ahoyhoy"  wrote in message

> > is there one constant rule that can be applied to any of the numbers to
> > arrive at the next number in the sequence?

> You then have T(n) for your sequence and can check it using the terms you
> have. After that, you can find any term by just subing in n

That's not what he's asking for, though. He wants a function which
applies to the number, not the index:

f(T_n) = T_(n+1)

Ian

"The Real Doctor"  wrote in message 
news:1191393966.481790.96540@y42g2000hsy.googlegroups.com...
> On 25 Sep, 20:42, "Brian Reay" <s...@website.invalid> wrote:
>> "Ahoyhoy"  wrote in message
>
>> > is there one constant rule that can be applied to any of the numbers to
>> > arrive at the next number in the sequence?
>
>> You then have T(n) for your sequence and can check it using the terms you
>> have. After that, you can find any term by just subing in n
>
> That's not what he's asking for, though. He wants a function which
> applies to the number, not the index:
>
> f(T_n) = T_(n+1)

True but I thought we could leave the next few steps to him (/her).

Hint: sub n+1 into  T(n) = An^2 + Bn + C, do some algebra, and you can 
relate T(n+1) to T(n)

Eitherway, having T(n) is no bad idea as you can check your working.

Now, if you've a better way to help the original poster, the stage is 
yours............

Brian

On 3 Oct, 22:35, "Brian Reay" <s...@website.invalid> wrote:

> Now, if you've a better way to help the original poster, the stage is
> yours............

What he asks for can't be done. What you suggest won't work. I'll
leave it to you to work out why ...

Ian

"Ian"  wrote in message 
news:1191448460.956345.176680@w3g2000hsg.googlegroups.com...
> On 3 Oct, 22:35, "Brian Reay" <s...@website.invalid> wrote:
>
>> Now, if you've a better way to help the original poster, the stage is
>> yours............
>
> What he asks for can't be done. What you suggest won't work. I'll
> leave it to you to work out why ...


T(n) = An^2 + Bn + C

You can find A, B, and C as I showed.

Look at the n+1 term:

T(n+1) = A(n+1) ^ 2 + B(n+1) + C

Expand and collect terms:

T(n+1) =   An^2+Bn + C  + 2A(n+1) + B

  An^2+Bn + C = T(n)

So:

T(n+1) = T(n) + 2A(n+1) + B

You know A & B, also T(n). Provided you know, or can find, n  you are home.

Not elegant but it can be done.

Brian

On 4 Oct, 17:25, "Brian Reay" <s...@website.invalid> wrote:
> "Ian"  wrote in message

> > What he asks for can't be done. What you suggest won't work. I'll
> > leave it to you to work out why ...

> You know A & B, also T(n). Provided you know, or can find, n  you are home.

But if you know n there is no need to go through this palaver. The
original question implied that you don't, and in that case the problem
is not solvable. Which is trivially simple to demonstrate - without
any algebra!

Ian

On 24 Sep, 20:39, "Ahoyhoy"  wrote:
> In the following number sequence the interval between each pair of adjacent
> numbers decreases by one as the sequences progresses.
>
> 2    10    17    23    28
>
> so...
>
>  2 >>> 10 = +8
> 10 >>> 17 = +7
> 17 >>> 23 = +6
> 23 >>> 28 = +5
>
> but ...
>
> is there one constant rule that can be applied to any of the numbers to
> arrive at the next number in the sequence?

Since debate has died down, I'll give away the answer.

No, you can't.

Write down the first seventeen terms of the series and see if you can
spot the problem.

Ian

Ian wrote:
> 
> On 24 Sep, 20:39, "Ahoyhoy"  wrote:
> > In the following number sequence the interval between each pair of adjacent
> > numbers decreases by one as the sequences progresses.
> >
> > 2    10    17    23    28
> >
> > so...
> >
> >  2 >>> 10 = +8
> > 10 >>> 17 = +7
> > 17 >>> 23 = +6
> > 23 >>> 28 = +5
> >
> > but ...
> >
> > is there one constant rule that can be applied to any of the numbers to
> > arrive at the next number in the sequence?
> 
> Since debate has died down, I'll give away the answer.
> 
> No, you can't.
> 
> Write down the first seventeen terms of the series and see if you can
> spot the problem.

Tell us more old chap!

-- 
Remove "antispam" and ".invalid" for e-mail address.
"He that giveth to the poor lendeth to the Lord, and shall be repaid," 
said Mrs Fairchild, hastily slipping a shilling into the poor woman's
hand.

On 12 Oct, 10:57, Frederick Williams <"Frederick
Williams"@antispamhotmail.co.uk.invalid> wrote:
> Ian wrote:
>
> > On 24 Sep, 20:39, "Ahoyhoy"  wrote:
> > > In the following number sequence the interval between each pair of adjacent
> > > numbers decreases by one as the sequences progresses.
>
> > > 2    10    17    23    28
>
> > > so...
>
> > >  2 >>> 10 = +8
> > > 10 >>> 17 = +7
> > > 17 >>> 23 = +6
> > > 23 >>> 28 = +5
>
> > > but ...
>
> > > is there one constant rule that can be applied to any of the numbers to
> > > arrive at the next number in the sequence?
>
> > Since debate has died down, I'll give away the answer.
>
> > No, you can't.
>
> > Write down the first seventeen terms of the series and see if you can
> > spot the problem.
>
> Tell us more old chap!

Did you try it?

Ian