Find the ${{t}_{n}}$ for the Arithmetic Progression where ${{t}_{3}}=22,\text{ }{{t}_{17}}=-20$.
(a) 3n-32
(b) 3n-31
(c) -3n+31
(d) -3n+32

Hint: First, we must know the following formula of the Arithmetic Progression to get the term ${{t}_{n}}$ which gives the $n^{th}$ term for the series of A.P as ${{t}_{n}}=a+\left( n-1 \right)d$. Then, by using this formula to get the two equations from the values that are given and solving them to get the values of a and d. Then, to get the general expression, substitute the values of a and d in the general expression of A.P which is ${{t}_{n}}=a+\left( n-1 \right)d$.

Complete step-by-step answer:
In this question, we are supposed to find the ${{t}_{n}}$ for the Arithmetic Progression where ${{t}_{3}}=22,\text{ }{{t}_{17}}=-20$.
So, before proceeding for this , we must know the following formula of the Arithmetic Progression to get the term ${{t}_{n}}$ which gives the $n^{th}$ term for the series of A.P.
Also, the terms are defined as d is the common difference between the consecutive terms of A.P and a is the first term of A.P.
Now, the general formula for the $n^{th}$ term of A. P is:
${{t}_{n}}=a+\left( n-1 \right)d$
So, by using this formula to get the two equations from the values that are given as:
$\begin{align}
  & {{t}_{3}}=a+\left( 3-1 \right)d \\
 & {{t}_{17}}=a+\left( 17-1 \right)d \\
\end{align}$
Now, we have the values of the above equations as ${{t}_{3}}=22,\text{ }{{t}_{17}}=-20$.
So, by using them we get the final equations as:
\[\begin{align}
  & 22=a+2d.....\left( i \right) \\
 & -20=a+16d.....\left( ii \right) \\
\end{align}\]
Now, by solving these two equations (i) and (ii), we can get the values of a and d.
So by subtracting equation (i) from equation (ii), we get:
\[-20-22=a+16d-a-2d\]
Now, b y solving the above expression for the value of d as:
$\begin{align}
  & 14d=-42 \\
 & \Rightarrow d=\dfrac{-42}{14} \\
 & \Rightarrow d=-3 \\
\end{align}$
So, we get the value of common difference as -3.
Now, substituting the value of d in equation (i), we get:
\[\begin{align}
  & 22=a+2\left( -3 \right) \\
 & \Rightarrow 22=a-6 \\
 & \Rightarrow a=22+6 \\
 & \Rightarrow a=28 \\
\end{align}\]
So, we get the value of the first term which is 28.
Now, to get the general expression, substitute the values of a and d in the general expression of A.P which is ${{t}_{n}}=a+\left( n-1 \right)d$.
So, by substituting the value of a=28 and d=-3, we get:
$\begin{align}
  & {{t}_{n}}=28+\left( n-1 \right)\left( -3 \right) \\
 & \Rightarrow {{t}_{n}}=28-3n+3 \\
 & \Rightarrow {{t}_{n}}=31-3n \\
 & \Rightarrow {{t}_{n}}=-3n+31 \\
\end{align}$
So, we get the value of ${{t}_{n}}$ as $-3n+31$.
Hence, option (c) is correct.

Note: Now, to solve these type of the questions we need to know the general expansion formula for the arithmetic progression as:
${{t}_{n}}=a+\left( n-1 \right)d$
Similarly, sometimes the question is asked about geometric progression whose formula is to be remembered where r is the common ratio as:
${{t}_{n}}=a{{r}^{n-1}}$