# The first and last term of an A.P. are a and l respectively.

Question:

The first and last term of an A.P. are $a$ and $/$ respectively. If $S$ is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2}-a^{2}}{k-(l+a)}$, then $k=$

Solution:

In the given problem, we are given the first, last term, sum and the common difference of an A.P.

We need to find the value of k

Here,

First term = a

Last term = l

Sum of all the terms = S

Common difference $(d)=\frac{l^{2}-a^{2}}{k-(l+a)}$

Now, as we know,

$l=a+(n-1) d$ .........(1)

Further, substituting (1) in the given equation, we get

$d=\frac{[a+(n-1) d]^{2}-a^{2}}{k-\{[a+(n-1) d]+a\}}$

$d=\frac{a^{2}+[(n-1) d]^{2}+2 a(n-1) d-a^{2}}{k-\{[a+(n-1) d]+a\}}$

$d=\frac{[(n-1) d]^{2}+2 a(n-1) d}{k-\{[a+(n-1) d]+a\}}$

Now, taking d in common, we get,

$d=\frac{[(n-1) d]^{2}+2 a(n-1) d}{k-\{[a+(n-1) d]+a\}}$

$1=\frac{(n-1)^{2} d+2 a(n-1)}{k-[2 a+(n-1) d]}$

$k-[2 a+(n-1) d]=(n-1)^{2} d+2 a(n-1)$

Taking (n-1) as common, we get,

$k-[2 a+(n-1) d]=(n-1)[(n-1) d+2 a]$

$k=n[(n-1) d+2 a]-[(n-1) d+2 a]+[2 a+(n-1) d]$

$k=n[(n-1) d+2 a]$

Further, multiplying and dividing the right hand side by 2, we get,

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

Now, as we know, $S=\frac{n}{2}[(n-1) d+2 a]$

Thus,

$k=2 S$

Therefore, the correct option is (b).