How do you find the \[{n^{th}}\] term rule for \[\dfrac{1}{2},1,\dfrac{3}{2},2.....\] ?

Hint: Here we are given with a series of numbers like. We have to find the nth term for the series given above. If we observe that the given series has a pattern as it is an arithmetic progression with a common difference of \[\dfrac{1}{2}\]. Thus this is our clue to lead. We know the formula for the \[{n^{th}}\] term of a series.

Complete step by step solution:
Given that the series is an A.P.
See the difference in second and third term is \[1 - \dfrac{1}{2} = \dfrac{1}{2}\]
Again the difference in the second and third term is \[\dfrac{3}{2} - 1 = \dfrac{1}{2}\]
Thus this is an A.P.
Now we know that \[{n^{th}}\] term of an A.P. is given by \[{a_n} = a + \left( {n - 1} \right)d\]
Where a is the first term of the series and d is the common difference. Also the term n is the number of terms in the series.
Here we have the values of a and d but not if n. so we will stop here only.
So, the correct answer is “\[{a_n} = a + \left( {n - 1} \right)d\]”.

Note: Here all you need to note is the series can either be an arithmetic progression or a geometric progression such that we know how to find the nth term of the series. If we know the number of terms then we can also find the nth term. Note that number of terms is a different concept than nth term.