MHB Rudra Rath's question at Yahoo Answers involving an arithmetic series

[MarkFL]

Here is the question:

An arithmetic progression question!?

the first and last terms of an A.P. are a and l respectively. if S be the sum of the terms, then show that the common difference is ( l^2 - a^2 ) / { 2S - (l+a) }

Here is a link to the question:

An arithmetic progression question!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Rudra Rath,

We are told:

(1) $$a_1=a$$

(2) $$a_n=a+(n-1)d=\ell$$

(3) $$S=\frac{n(a+\ell)}{2}$$

Now, solving (2) for $d$, we find:

(4) $$d=\frac{\ell-a}{n-1}$$

Solving (3) for $n$, we find:

(5) $$n=\frac{2S}{a+\ell}$$

Substituting for $n$ from (5) into (4), we obtain:

$$d=\frac{\ell-a}{\frac{2S}{a+\ell}-1}\cdot\frac{\ell+a}{\ell+a}=\frac{\ell^2-a^2}{2S-(\ell+a)}$$

Shown as desired.

To Rudra Rath and any other guests viewing this topic, I invite and encourage you to post other arithmetic series questions in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.

 

[Prove It]

An alternative method:

If you have an arithmetic series, the sum of the first n terms can be found either using [math]\displaystyle S_n = \frac{n}{2} \left( a+ l \right) [/math] or [math]\displaystyle S = \frac{n}{2} \left[ 2a + \left( n - 1 \right) d \right] [/math]. So clearly we can see

[math]\displaystyle \begin{align*} \frac{n}{2} \left( a + l \right) &= \frac{n}{2} \left[ 2a + \left( n - 1 \right) d \right] \\ a + l &= 2a + \left( n - 1 \right) d \\ l - a &= \left( n - 1 \right) d \\ \frac{l - a}{n - 1} &= d \\ \frac{ \left( l - a \right) \left( l + a \right) }{ \left( a + l \right) \left( n - 1 \right) } &= d \\ \frac{ l^2 - a^2}{ n \left( a + l \right) - \left(a + l \right) } &= d \\ \frac{ l^2 - a^2 }{ 2 \left[ \frac{n}{2} (a + l) \right] - (a + l) } &= d \\ \frac{ l^2 - a^2 }{ 2S_n - (a + l) } &= d \end{align*}[/math]