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**Hint:**In the question the first and the last terms of an AP are given also it is given that the AP has \[60\] terms. Using this we can find out the common difference between two consecutive terms and then find out the \[{27^{th}}\] term by using the formula of ${n^{th}}$ term in an AP.

**Complete step-by-step answer:**The first term of the AP is given as \[6\] and the last term is \[124\].

In any standard AP, let us assume $a$ to be the first term and $d$ be the common difference of the AP.

As the common difference $d$ is not given in the question, we will have to find that using the formula of ${n^{th}}$ term of an AP.

The ${n^{th}}$ term of an AP is given by the formula $t = a + (n - 1)d$, where \[a\] is the first term of AP, \[d\] is the common difference and \[t\] is the term itself.

The first term of the AP is \[6\] and the last term of the AP is \[124\] which is the \[{60^{th}}\] term as mentioned in the question.

Substituting these values in the formula of ${n^{th}}$ term, we get

\[

\Rightarrow t = a + (n - 1)d \\

\Rightarrow 124 = 6 + (60 - 1)d \\

\Rightarrow 124 = 6 + 59d \\

\Rightarrow 59d = 124 - 6 \\

\Rightarrow 59d = 118 \\

\Rightarrow d = 2 \\

\]

So, we get the value of \[d\] as \[2\].

Now, we again substitute the values of \[a\], \[d\] and \[n\] as \[6\], \[2\] and \[27\] respectively to get the \[{27^{th}}\] term of the AP.

Thus, on substituting, we get

\[

\Rightarrow t = a + (n - 1)d \\

\Rightarrow t = 6 + (27 - 1)2 \\

\Rightarrow t = 6 + (26)2 \\

\Rightarrow t = 6 + 52 \\

\Rightarrow t = 58 \\

\]

**Thus, the \[{27^{th}}\] term of the AP will be \[58\].**

**Note:**For finding any term of an AP, we need the first term and the common difference. In the question, only the first term was given, so we had to find the common difference by using the data about the last term. For solving questions of AP, students must be well-versed with the formulas to solve the question much faster, otherwise if we solve them by adding the difference many times and then getting the term will be a lot of time wastage.

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