Write the first term, common difference and ${{n}^{th}}$ term of AP.
(a)-2, 2, 6, 10………
(b)$\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3},\dfrac{13}{3},.......$

Hint: We will write the formula for ${{n}^{th}}$ term of AP, after that we know the value of first term of AP and we can find the common difference by subtracting the two terms of the AP and then we can use these values in the formula for ${{n}^{th}}$ term of AP.

Complete step-by-step answer:
Now we will start writing the solution by first finding all the values needed to find the ${{n}^{th}}$ term of AP.
For (a): -2, 2, 6, 10………
We get the first term as a = -2,
We get the common difference (d) = $2-\left( -2 \right)=4$
The formula for ${{n}^{th}}$ term of AP is: $a+\left( n-1 \right)d$ . Therefore, substituting the known terms, we get
$\begin{align}
  & =-2+\left( n-1 \right)4 \\
 & =4n-6 \\
\end{align}$
For (b): $\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3},\dfrac{13}{3},.......$
We get the first term as a = $\dfrac{1}{3}$ ,
We get the common difference (d) = $\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}$
The formula for ${{n}^{th}}$ term of AP is: $a+\left( n-1 \right)d$ . Therefore, substituting the known terms, we get
$\begin{align}
  & =\dfrac{1}{3}+\dfrac{4\left( n-1 \right)}{3} \\
 & =\dfrac{4n}{3}-\dfrac{4}{3}+\dfrac{1}{3} \\
 & =\dfrac{4n}{3}-1 \\
\end{align}$
Now we have found all the required values that were asked in the question.

Note: The first term of the AP will be the same, we can take any two consecutive terms of an AP and then subtract it to find the value of common difference, here we have taken the first two terms to find out the value of the common difference. In the second part the terms are in fraction so one should be careful and while finding the common difference as it might be a bit confusing.