How do you find \[{{a}_{1}}\] in the arithmetic series with \[{{S}_{7}}=287\] and \[d=12\]?

Hint: In order to find the solution of this question, we should know that sum of terms of an arithmetic progression is given by \[{{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)\] where \[{{a}_{1}}\] represents the first term of progression, \[d\] represents the common difference of progression, \[n\] represents the nth term of progression and \[{{S}_{n}}\] represents the sum of first n terms of progression. By using this formula, we will get our answer.

Complete step by step answer:
According to the question, we have been asked to find the value of \[{{a}_{1}}\] in the arithmetic series with \[{{S}_{7}}=287\] and \[d=12\].
Now, we know that arithmetic progression is a series of n numbers such that the difference between two consecutive numbers remains constant.
Also, we know that for arithmetic progression with first term as \[{{a}_{1}}\] and common difference \[d\] then sum of n terms is given by \[{{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)\].
Now, according to the question, we have been given that \[{{S}_{7}}=287\] and \[d=12\]. Therefore, for n = 7, we can write \[{{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)\] as
\[{{S}_{7}}=\dfrac{7}{2}\left( 2{{a}_{1}}+\left( 7-1 \right)12 \right)\]
Now, we will put the value \[{{S}_{7}}=287\], so, we get
\[\Rightarrow 287=\dfrac{7}{2}\left( 2{{a}_{1}}+\left( 7-1 \right)12 \right)\]
Now, we will simplify the obtained expression to get our answer. Therefore, we can write
\[\Rightarrow 287=\dfrac{7}{2}\left( 2{{a}_{1}}+\left( 6 \right)12 \right)\]
Now, we will open the inner brackets and make necessary calculations. Therefore, we get
\[\Rightarrow 287=\dfrac{7}{2}\left( 2{{a}_{1}}+6\times 12 \right)\]
\[\Rightarrow 287=\dfrac{7}{2}\left( 2{{a}_{1}}+72 \right)\]
Now, we will cross multiply the equation, therefore, we get
\[\Rightarrow 287\times 2=7\left( 2{{a}_{1}}+72 \right)\]
On further simplifications, we get
\[\Rightarrow 574=7\left( 2{{a}_{1}}+72 \right)\]
Now, we will divide both sides of equation by 7, therefore, we get
\[\Rightarrow \dfrac{574}{7}=\dfrac{7\left( 2{{a}_{1}}+72 \right)}{7}\]
And we can also write it as
\[\Rightarrow 82=\left( 2{{a}_{1}}+72 \right)\]
Now, we will subtract 72 from both sides of the equation. Therefore, we get
\[\Rightarrow 82-72=2{{a}_{1}}+72-72\]
\[\Rightarrow 10=2{{a}_{1}}\]
Now, we will divide both sides by 2. Therefore, we get
\[\Rightarrow \dfrac{10}{2}=\dfrac{2{{a}_{1}}}{2}\]
And we can also write it as
\[\therefore {{a}_{1}}=5\]
Thus, the first term of this arithmetic progression is \[{{a}_{1}}=5\].

Note:
Many a times, students get confused in the formula of sum of n terms, that is \[{{S}_{n}}=\dfrac{n}{2}(2{{a}_{1}}+(n-1)d)\] and nth term of progression, that is \[{{a}_{n}}={{a}_{1}}+(n-1)d\]. One should keep the formulas in their mind and also be able to derive them quickly. Other formula for sum of n terms is \[{{S}_{n}}=\dfrac{n}{2}({{a}_{1}}+{{a}_{n}})\] where \[{{a}_{n}}\]is the nth term of the arithmetic progression. Also, we have to be very careful while doing calculations to avoid calculation mistakes.