In an \[AP\], given ${a_{12}} = 37,\,d = 3, $find$ a\,{\text{and}}\,{S_{12}}.$

Hint: Here before solving this question we need to know the following formula:-
The ${n^{th}}$term of an Arithmetic progression or \[A.P.\]is given by,
${a_n} = a + \left( {n - 1} \right)d\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$
The sum of ${n^{th}}$term of an Arithmetic progression or \[A.P.\]is given by,
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\,\,\,\,\,\,\, \ldots \left( 2 \right)$
Where,
\[
  {a_n} = {n^{{\text{th}}}}{\text{ term of an AP}} \\
  a = {\text{ First term}} \\
  d = {\text{ Common Difference of an AP}} \\
 \]

Complete step-by-step answer:
According to this question we have,
${a_{12}} = 37\,{\text{and }}d = 3$
We can write the given values as,
\[
  {a_{12}} = 37 \\
  n = 12 \\
  a = ? \\
  d = {\text{ 3}} \\
 \]
We substitute all the values in equation \[\left( 1 \right).\]
$
  37 = a + \left( {12 - 1} \right)3 \\
  37 = a + \left( {11} \right)3 \\
  37 = a + 33 \\
  a = 4 \\
$
In the second part, we need to calculate the value${S_{12}}$.
So, we can write it as
\[
  {S_{12}} = ? \\
  n = 12 \\
  a = 4 \\
  d = {\text{ 3}} \\
\]
Substitute all the value in equation \[\left( 2 \right)\]
$
  {S_{12}} = \dfrac{{12}}{2}\left( {2\left( 4 \right) + \left( {12 - 1} \right)3} \right) \\
   = 6\left( {8 + 33} \right) \\
$
Further solving
$
   = 6 \times 41 \\
   = 246 \\
$
Thus, the value of $a = 4$ and ${S_{12}} = 246$.

Note: While writing the Solution do not get confused with the formula ${a_n}\,\,{\text{and}}\,\,{S_n}$.
The value in the lower subscript denotes the number of terms.
It is advisable to write all the values and then choose the formula according to the need and the requirement of the question.