Answer
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Hint- This question is solved by using the formula for ${n^{th}}$ term when a series is in AP.
Now,
Given that $ - 5,K, - 1$ are in AP.
And we have to find the value of $k$ .
Now we know the formula to find the ${n^{th}}$ of an AP
${a_n} = {a_1} + \left( {n - 1} \right)d$
Here ${a_n}$ is the ${n^{th}}$ term,
${a_1}$ is the first term,
$d$ is the common difference and
$n$ is the number of terms which are to be found.
Now,
${a_3} = {a_1} + \left( {n - 1} \right)d$
Here, ${a_3} = - 1$ , ${a_1} = - 5$, $n = 3$ and
$
d = k - \left( { - 5} \right) \\
= k + 5 \\
$
Putting the value of these we get,
$
- 1 = - 5 + \left( {3 - 1} \right)\left( {k + 5} \right) \\
{\text{or }} - 1 = - 5 + 2\left( {k + 5} \right) \\
{\text{or }} - 1 + 5 = 2k + 10 \\
{\text{or }}4 = 2k + 10 \\
{\text{or }}4 - 10 = 2k \\
{\text{or }} - 6 = 2k \\
{\text{or }}k = - 3 \\
$
Thus, the correct option is $\left( B \right)$.
Note- Whenever we face such types of questions the key concept is that we should know the formulas when a series is in AP. Like we did in this question here, we apply the formula for ${n^{th}}$ term and we find the solution.
Now,
Given that $ - 5,K, - 1$ are in AP.
And we have to find the value of $k$ .
Now we know the formula to find the ${n^{th}}$ of an AP
${a_n} = {a_1} + \left( {n - 1} \right)d$
Here ${a_n}$ is the ${n^{th}}$ term,
${a_1}$ is the first term,
$d$ is the common difference and
$n$ is the number of terms which are to be found.
Now,
${a_3} = {a_1} + \left( {n - 1} \right)d$
Here, ${a_3} = - 1$ , ${a_1} = - 5$, $n = 3$ and
$
d = k - \left( { - 5} \right) \\
= k + 5 \\
$
Putting the value of these we get,
$
- 1 = - 5 + \left( {3 - 1} \right)\left( {k + 5} \right) \\
{\text{or }} - 1 = - 5 + 2\left( {k + 5} \right) \\
{\text{or }} - 1 + 5 = 2k + 10 \\
{\text{or }}4 = 2k + 10 \\
{\text{or }}4 - 10 = 2k \\
{\text{or }} - 6 = 2k \\
{\text{or }}k = - 3 \\
$
Thus, the correct option is $\left( B \right)$.
Note- Whenever we face such types of questions the key concept is that we should know the formulas when a series is in AP. Like we did in this question here, we apply the formula for ${n^{th}}$ term and we find the solution.
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