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 $k=$

(a) S

(b) 2S

(c) 3S

(d) none of these

Solution:

(b) 2S

Given:

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

$\Rightarrow(l+a)=\frac{2 S}{n}$

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

$\Rightarrow d=\frac{(l+a)(l-a)}{k-(l+a)}$

$\Rightarrow d=\frac{[(n-1) d] \times \frac{2 S}{n}}{k-\frac{25}{n}}$

$\Rightarrow k-\frac{2 S}{n}=(n-1) \frac{2 S}{n}$

$\Rightarrow k=\frac{2 S}{n}(n-1+1)$

$\Rightarrow k=2 S$

 

 

 

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