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

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2 s

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3 s

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None of these

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Solution

The correct option is B $2 s$
We have, $S = \frac{n}{2} (a + l) \Rightarrow \frac{2 S}{a + l} = n$ ...(1)

Also, $l = a + (n - 1) d \Rightarrow d = \frac{l - a}{n - 1} = \frac{l - a}{\frac{2 S}{a + l} - 1}$ (from (1))
$= \frac{l^{2} - a^{2}}{2 S - (l + a)}$

$∴ k = 2 S$

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Similar questions

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 given by $\frac{l^{2} - a^{2}}{k - (l + a)}$ , then k =

A. S

B 2S

C 3S

D None of these

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\frac{l^{2} - a^{2}}{k - (l + a)}

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(b) 2S
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\frac{(l^{2} - a^{2})}{2 S - a - l}

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a

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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−a2k−(l+a), then k is equal to

The correct option is B 2sWe have, S=n2(a+l)⇒2Sa+l=n ...(1)Also, l=a+(n−1)d⇒d=l−an−1=l−a2Sa+l−1 (from (1))=l2−a22S−(l+a)∴k=2S

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

The correct option is B $2 s$
We have, $S = \frac{n}{2} (a + l) \Rightarrow \frac{2 S}{a + l} = n$ ...(1)

Also, $l = a + (n - 1) d \Rightarrow d = \frac{l - a}{n - 1} = \frac{l - a}{\frac{2 S}{a + l} - 1}$ (from (1))
$= \frac{l^{2} - a^{2}}{2 S - (l + a)}$

$∴ k = 2 S$