If \[a\] is the first term, 1 is the common difference and \[b\] is the last term of an AP, then its sum is.
A.\[\dfrac{{\left( {a + b} \right)\left( {1 + a - b} \right)}}{2}\]
B.\[\dfrac{{\left( {a + b} \right)\left( {1 - a + b} \right)}}{2}\]
C.\[\dfrac{{\left( {a + b} \right)\left( {1 - a} \right)}}{2}\]
D.\[\left( {a + b} \right)\left( {1 - a + b} \right)\]

Hint: Here we will first assume the number of terms in an AP to be any variable. Then we will use the formula of the general term of an AP to find the required number of terms present in an AP. At last, we will use the formula of the sum of \[n\] terms of an AP to get the required answer.

Formula Used:
We will use the following formulas:
1.The sum of the terms of an AP is given by \[{\rm{Sum}} = \dfrac{n}{2}\left( {a + l} \right)\] where, \[n\] is the number of terms, \[l\] is the last term of an AP and \[n\] is the number of terms of an AP.
2.The formula of the general term of an AP is given by \[{T_n} = a + \left( {n - 1} \right)d\].

Complete step-by-step answer:
It is given that \[a\] is the first term, 1 is the common difference and \[b\] is the last term of an AP.
Let the number of terms in an AP to be \[n\].
We know the value of the last term. So we will use the formula of the general term of an AP for the last term of an AP.
Now, we will substitute the value of the first term, last term, number of terms and the value of the common difference in the formula \[{T_n} = a + \left( {n - 1} \right)d\]. Therefore, we get
\[b = a + \left( {n - 1} \right) \times 1\]
On further simplification, we get
\[ \Rightarrow b = a + n - 1\]
Now, adding 1 to both sides, we get
\[\begin{array}{l} \Rightarrow b + 1 = a + n - 1 + 1\\ \Rightarrow b + 1 = a + n\end{array}\]
Now, we will subtract the term \[a\] from both sides. So, we get
\[\begin{array}{l} \Rightarrow b + 1 - a = a + n - a\\ \Rightarrow b + 1 - a = n\end{array}\]
\[ \Rightarrow n = b - a + 1\] …………. \[\left( 1 \right)\]
Now we will find the sum of \[n\] terms.
Using the formula \[{\rm{Sum}} = \dfrac{n}{2}\left( {a + l} \right)\] for the given AP, we get
\[{\rm{Sum}} = \dfrac{n}{2}\left( {a + b} \right)\]
Now, we will substitute the value of the number of terms of an AP from equation \[\left( 1 \right)\].
\[ \Rightarrow {\rm{Sum}} = \dfrac{{\left( {1 - a + b} \right)}}{2}\left( {a + b} \right)\]
Hence, the correct option is option B.

Note: Here we have obtained the value of the sum of terms of an AP. Here, AP stands for the Arithmetic Progression and it is defined as the sequence in which the difference between any two consecutive terms of an arithmetic progression is constant. If we add any constant term to each term of an AP, then the resultant sequence will also be in AP. A real-life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.