How do you give a recursive formula for the arithmetic sequence where the \[4th\] term is\[\;3\] ; \[20th\] term is \[35\]?

Hint: We have given the arithmetic sequence where the \[4th\] term is\[\;3\] ; \[20th\] term is \[35\]. We know that general formula for arithmetic progression is given as \[{a_n} = {a_1} + (n - 1) \cdot d\] , where ‘an’ is representing the $nth$ term of the arithmetic series, ‘a1 ‘ is representing the very first term of the arithmetic series, and ‘d’ is representing the common difference. We will make equations from the given constraints then simplify them and will get the required result.

Complete step by step solution:
We have given the arithmetic sequence where the \[4th\] term is\[\;3\] ; \[20th\] term is \[35\] ,
We know that general formula for arithmetic progression is given as \[{a_n} = {a_1} + (n - 1) \cdot d\] ,
 where ‘an’ is representing the $nth$ term of the arithmetic series, ‘a1 ‘ is representing the very first term of the arithmetic series, and ‘d’ is representing the common difference.
According to the question ,
We will get ,
\[
  {a_4} = {a_1} + (4 - 1) \cdot d \\
  3 = {a_1} + 3d...............(1) \\
 \]
And
\[
  {a_{20}} = {a_1} + (20 - 1) \cdot d \\
  35 = {a_1} + 19d...............(2) \\
 \]
After solving $(1),(2)$ , we will get ,
$
  d = 2 \\
  {a_1} = - 3 \\
 $
With the above data we can write the recursive formula for the arithmetic sequence where the \[4th\] term is\[\;3\] ; \[20th\] term is \[35\] as
\[
  {a_n} = - 3 + (n - 1) \cdot (2) \\
  {a_n} = 2n - 5 \\
 \]

Note: While solving $(1),(2)$ we have to substitute the expression from $(1)$ for $d$ that is $d = \dfrac{{3 - {a_1}}}{3}$into the given second equation. : Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.