A.P

Posted August 3rd, 2008 by alpha

The first and the last term of an A.P are a and l respectively.If S is the sum of all the terms of A.P and the common difference is given by

$\frac{l^2-a^2}{kS-(l+a)}=d$

, then k=2

Show me all the steps please.

Comments

On August 3rd, 2008 Structure says:

You have missed in your initial post to write kS in your fraction

$d=\frac{l^2-a^2}{kS-(l+a)}$

$S=s_n=\frac{n(a_1+a_n)}{2}$

$2s_n=2S=n(a_1+a_n)=n(a+l)$

Now you may write

$l^2-a^2=d(kS-(l+a))$

$(l-a)(l+a)+d(l+a)=(l+a)(l-a+d)=\frac{2s_n}{n}nd=2s_nd=kdS=kds_n$

so k=2

On August 4th, 2008 alpha says:

My question does not contain S in the denominator.

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On August 4th, 2008 Structure says:

You have the answer
In this case your constant is 2S