The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is (l2-a2/k-(l+a)), then k is equal to

Question

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is (l2-a2/k-(l+a)), then k is equal to (A) S (B) 2 S (C) 3 S (D) None of these

Tardigrade Admin · Accepted Answer

We have, S=(n/2)(a+l) ⇒ (2 S/a+l)=n (1) Also, l=a+(n-1) d ⇒ d=(l-a/n-1) =(l-a/(2 S)a+l-1)[ Using (1) ] =(l2-a2/2 S-(l+a)) ∴ k=2 S .

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

S

2 S

3 S

None of these

We have, $S = \frac{n}{2} (a + l)$
$\Rightarrow \frac{2 S}{a + l} = n (1)$
Also, $l = a + (n - 1) d$
$\Rightarrow d = \frac{l - a}{n - 1}$
$= \frac{l - a}{\frac{2 S}{a + l} - 1} [$ Using (1) $]$
$= \frac{l ^{2} - a ^{2}}{2 S - ( l + a )}$
$∴ k = 2 S .$

Q. The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is k−(l+a)l2−a2​, then k is equal to

S

2 S

3 S

None of these

We have, S=2n​(a+l) ⇒a+l2S​=n(1) Also, l=a+(n−1)d ⇒d=n−1l−a​ =a+l2S​−1l−a​[ Using (1) ] =2S−(l+a)l2−a2​ ∴k=2S.

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

We have, $S = \frac{n}{2} (a + l)$
$\Rightarrow \frac{2 S}{a + l} = n (1)$
Also, $l = a + (n - 1) d$
$\Rightarrow d = \frac{l - a}{n - 1}$
$= \frac{l - a}{\frac{2 S}{a + l} - 1} [$ Using (1) $]$
$= \frac{l ^{2} - a ^{2}}{2 S - ( l + a )}$
$∴ k = 2 S .$