Arithmetic progression question

**MichaelLight** · March 11th 2011, 06:33 PM

Series Q is an arithmetic series such that the sum of its first n even terms is more than the sum of its first n odd terms by 4n. Find the common difference of the series Q. The answer provided is 4.

Can you help me? Thanks.

**Prove It** · March 11th 2011, 06:37 PM

I don't quite understand the question as written.

Are you saying that $\displaystyle t_2 + t_4 + t_6 + \dots = 4n + t_1 + t_3 + t_5 + \dots$ ?

**MichaelLight** · March 11th 2011, 06:40 PM

Originally Posted by Prove It

I don't quite understand the question as written.

Are you saying that $\displaystyle t_2 + t_4 + t_6 + \dots = 4n + t_1 + t_3 + t_5 + \dots$ ?

I guess that's what the question means...

**Prove It** · March 11th 2011, 06:50 PM

In that case you will need to consider two cases.

Case 1: Your series has an odd number of terms, which means $\displaystyle t_n$ will be on the RHS.

Case 2: Your series has an even number of terms, which means $\displaystyle t_n$ will be on the LHS.

Let's look at Case 1:

$\displaystyle t_2 + t_4 + t_6 + \dots + t_{n-1} = 4n + t_1 + t_3 + t_5 + \dots + t_n$ .

(Excluding the $\displaystyle 4n$ term), each side of the equation is an arithmetic series.

Since there were an odd number of terms in your original series, that means there are $\displaystyle \frac{n-1}{2}$ terms on the LHS, and $\displaystyle \frac{n + 1}{2}$ terms on the RHS.

Can you sum these two smaller arithmetic series?

**MichaelLight** · March 11th 2011, 07:07 PM

Originally Posted by Prove It

Since there were an odd number of terms in your original series, that means there are $\displaystyle \frac{n-1}{2}$ terms on the LHS, and $\displaystyle \frac{n + 1}{2}$ terms on the RHS.

Can you sum these two smaller arithmetic series?

Sorry but please allow me to ask, why $\displaystyle \frac{n-1}{2}$ terms on the LHS and $\displaystyle \frac{n + 1}{2}$ terms on the RHS (how do you know it)? and what do you mean by 'sum these two smaller arithmetic series'? Please enlighten me...

P/S: I have difficulty in forming an appropriate equation to find the common difference... can you help me..?

**Prove It** · March 11th 2011, 11:36 PM

If there are an odd number of terms, that means on the LHS there will be one less term than on the right.

How many terms are there on each side? Well, if you halve an odd number, you'll get a remainder of $\displaystyle \frac{1}{2}$ . This extra half will have to go to the right hand side.

That means the LHS will have $\displaystyle \frac{n}{2} - \frac{1}{2} = \frac{n-1}{2}$ terms, and the RHS will have $\displaystyle \frac{n}{2} + \frac{1}{2} = \frac{n+1}{2}$ terms.

Also, do you see that each term on the LHS will have its own common difference, which is twice the common difference of your original series? Same on the RHS. That means they are both arithmetic series...

How do you sum an arithmetic series?

**MichaelLight** · March 12th 2011, 01:45 AM

Originally Posted by Prove It

If there are an odd number of terms, that means on the LHS there will be one less term than on the right.

How many terms are there on each side? Well, if you halve an odd number, you'll get a remainder of $\displaystyle \frac{1}{2}$ . This extra half will have to go to the right hand side.

That means the LHS will have $\displaystyle \frac{n}{2} - \frac{1}{2} = \frac{n-1}{2}$ terms, and the RHS will have $\displaystyle \frac{n}{2} + \frac{1}{2} = \frac{n+1}{2}$ terms.

Ok, i think i understand this part now.

Originally Posted by Prove It

Also, do you see that each term on the LHS will have its own common difference, which is twice the common difference of your original series? Same on the RHS. That means they are both arithmetic series...

Yes i see it...

Originally Posted by Prove It

How do you sum an arithmetic series?

Just apply this formula

, isn't it?

Then, how should i continue now....?

**Prove It** · March 12th 2011, 01:58 AM

OK, so that we don't get confused, for our original series we'll use lowercase letters, and for each of the smaller series I'll use capitals...

You're saying that to sum each arithmetic series, use the formula $\displaystyle S_N = \frac{N}{2}[2A + (N-1)D]$ .

Now, the difference $\displaystyle D = 2d$ , where $\displaystyle d$ is the difference from the original series.

Working with the LHS... We have already established that $\displaystyle N = \frac{n - 1}{2}$ , and we should note that $\displaystyle A = t_2 = a + d$ .

So $\displaystyle LHS = \frac{N}{2}[2A + (N-1)D]$

$\displaystyle = \frac{\frac{n-1}{2}}{2}\left[2(a+d) + \left(\frac{n-1}{2}\right)(2d)\right]$

$\displaystyle = \frac{n-1}{4}\left[2a + 2d + (n-1)d\right]$

$\displaystyle = \frac{n-1}{4}\left[2a + (n + 1)d\right]$ .

Now try to arrive at a similar expression for the RHS.

**MichaelLight** · March 12th 2011, 02:26 AM

Sorry but do we have something wrong with this step? $\displaystyle = \frac{\frac{n-1}{2}}{2}\left[2(a+d) + \left(\frac{n-1}{2}\right)(2d)\right]$ Cause i get different answer...

For the RHS i get $Arithmetic progression question-msp815819efdh1f0agdg09700006831i1ad58a5ch6g.gif$ ... Please correct me if i am wrong...

**Prove It** · March 12th 2011, 02:52 AM

Yes, you're right about there being a mistake - it should be $\displaystyle \frac{n-1}{4}\left[2(a + d) + \left(\frac{n-1}{2} - 1\right)(2d)\right]$ , which you can simplify further.

I agree with your evaluation of the series on the RHS, don't forget to add $\displaystyle 4n$ though...

**MichaelLight** · March 12th 2011, 03:08 AM

Solved it. Thanks for your time and patient Prove it. ^^

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