# Math Help - Arithmetic Progression

1. ## Arithmetic Progression

2 Questions:

1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.

2. Hi

I don't know what the meaning of "special" in "special arithmetic progression is...
Originally Posted by Tangera
2 Questions:

1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.
Let r be the constant progression of this sequence.

We have $731=663+2r$ because they're separated by 2 terms.

So $r=34$

Now, note that $731=34*21+17$
It all depends on where you want to start the sequence, but if it has to be the first positive term, it would be $a_0=17$ and $M=21$

With these informations, I guess you can continue, can't you ?

3. Originally Posted by Moo
Hi

I don't know what the meaning of "special" in "special arithmetic progression is...

Let r be the constant progression of this sequence.

We have $731=663+2r$ because they're separated by 2 terms.

So $r=34$

Now, note that $731=34*21+17$
It all depends on where you want to start the sequence, but if it has to be the first positive term, it would be $a_0=17$ and $M=34$ Shouldn't it be M = 21 ?

With these informations, I guess you can continue, can't you ?
..

4. Thank you for your help! I would like to know if it is possible to work out that M = 22 without using the fact that 731 = 34 x 21 + 17...because that would involve a bit of trial and error right?

Could you guide me to solve question 2 as well? Thank you!

2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.

5. Hi

Thank you for your help! I would like to know if it is possible to work out that M = 22 without using the fact that 731 = 34 x 21 + 17...because that would involve a bit of trial and error right?
Not trial and error. It's just that you should know this :

$a_n=a_m+(n-m)r$

Here, n=M and m=0
And n-m=21, because r=34.

So M-0=21 --> M=21

I'm sorry for the second problem, I can't really see it for the moment

6. Sorry! I still don't understand why n-m=21 because r=34...? Aren't there 2 variables in that same equation??

7. Originally Posted by Tangera
Sorry! I still don't understand why n-m=21 because r=34...? Aren't there 2 variables in that same equation??

Hmmm... "because" was written because I sort of identified the numbers. To resume :

$a_n=a_m+(n-m)r$

Here, we have : $\underbrace{a_M}_{731}=\underbrace{a_0}_{17}+34 \times 21$

So it's clear that $a_M=a_n$, that is to say n=M
and $a_m=a_0$, that is to say m=0

There remain (n-m) and r to determine. r is the progression, so 34=r. Thus remains (n-m), which will be 21.
we know from above that n-m=M-0.
So M-0=21 ----> M=21

I hope this clears out the problem...sorry for not being clear

8. Originally Posted by Tangera
2 Questions:

1) The (M-2)th term of a special arithmetic progression is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

2) The sum of the first N terms of another special arithmetic progression is S. Find, in terms of S, the sum of the next N terms.

To prove the data is insufficient for a fixed N, note that the A.P with a = 1 and d=2 gives $N^2$ as the sum of first N terms and thereby the sum of next N terms is $3N^2$.
However The A.P with a = N, d=0 gives the same sum of first N terms but the sum of next N terms is $N^2$.
All I can prove is Sum of Next N terms $= S + n^2d$.