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**Hint:**Here, we will find the common difference and then use the formula term of \[n\]th term of the arithmetic progression \[{T_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference. Apply this formula, and then substitute the value of \[a\],\[d\] and \[n\] in the obtained equation to find the required A.P and then finally calculate the sum using the formula of sum of \[n\]th term of the arithmetic progression A.P., that is,\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference.

**Complete step-by-step answer:**We are given that the first, second terms of an A.P. are 27, 24 respectively.

Since it is given that the middle form is 0 and for the middle term to exist we should have odd.

Finding the common difference \[d\] by subtracting 27 from 24, we get

\[

\Rightarrow d = 24 - 27 \\

\Rightarrow d = - 3 \\

\]

Using the formula of \[n\]th term of the arithmetic progression A.P., that is,\[{T_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference, we get

\[

\Rightarrow {T_2} = a + \left( {2 - 1} \right)d \\

\Rightarrow {T_2} = a + d \\

\]

Substituting the value of \[{T_2}\] in the left hand side of the above equation, we get

\[ \Rightarrow 24 = a + d\]

Using the formula of sum of \[n\]th term of the arithmetic progression A.P., that is,\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference, we get

\[

\Rightarrow - 30 = \dfrac{n}{2}\left( {2\left( {27} \right) + \left( {n - 1} \right)\left( { - 3} \right)} \right) \\

\Rightarrow - 30 = \dfrac{n}{2}\left( {54 - 3n + 3} \right) \\

\Rightarrow - 30 = \dfrac{n}{2}\left( {57 - 3n} \right) \\

\]

Cross-multiplying the above equation and then rearrange the terms, we get

\[

\Rightarrow - 60 = n\left( {57 - 3n} \right) \\

\Rightarrow - 60 = 57n - 3{n^2} \\

\Rightarrow 3{n^2} - 57n - 60 = 0 \\

\]

Splitting the middle term in the left hand side of the above equation, we get

\[

\Rightarrow 3{n^2} - 60n + 3n - 60 = 0 \\

\Rightarrow 3n\left( {n - 20} \right) + 3\left( {n - 20} \right) = 0 \\

\Rightarrow \left( {3n + 3} \right)\left( {n - 20} \right) = 0 \\

\Rightarrow n = 20, - 1 \\

\]

Since the negative value is discarded, so the required value is 20.

**Hence, option B is correct.**

**Note:**In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula of sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term, but have the to find the value of \[{a_n}\] , so it will be wrong. We can also find the value of \[n\]th term by find the value of \[{S_n} - {S_{n - 1}}\], where \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the \[{a_n}\] is the \[n\]th term in the arithmetic progression.

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