Question

Find the value of ‘a’ and \[{{S}_{12}}\] if \[{{a}_{12}}=37,\ d=3\] .
(where ‘a’ is the first term and ‘d’ is the common difference of the arithmetic progression)

Answer 1

Hint: The formula for writing \[{{n}^{th}}\] term of an arithmetic progression is
\[{{n}^{th}}\ term=a+(n-1)d\] (where ‘a’ is the first term and ‘d’ is the common difference of the arithmetic progression)
The formula for writing the sum of first n terms of an arithmetic progression is
 \[{{S}_{n}}=\dfrac{n}{2}\]\[(2a+(n-1)d)\]

Complete step-by-step answer:
As mentioned in the question, it is given that the common difference of the given arithmetic progression is 3, therefore d is equal to 3. It is also given in the question that the \[{{12}^{th}}\] term of the arithmetic progression is 37.
\[{{a}_{12}}=37\]
Using the formula for writing the \[{{n}^{th}}\] term of an arithmetic progression, we will write the formula for 12th term as follows
\[\begin{align}
  & =a+\left( 12-1 \right)d \\
 & =a+11\times 3 \\
 & =a+33 \\
\end{align}\]
And as we know this value is equal to 37, hence, the equation can be formed as
 \[\begin{align}
  & a+33=37 \\
 & a=4 \\
\end{align}\]
Now, we have the value of the first term, therefore, we can find the value of the sum of the first 12 terms of the arithmetic progression using the sum formula as mentioned in the hint as follows
\[\begin{align}
  & {{S}_{12}}=\dfrac{12}{2}\left( 2a+(12-1)\times 3 \right) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{12}{2}\left( 2\times 4+11\times 3 \right) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =6(8+33) \\
 & \ \ \ \ \ \ \ \ \ \ \ \ =246 \\
\end{align}\]
Hence, the value of the first term i.e. a is 4 and the value of the sum of first 12 terms of the arithmetic progression is 246.

Note: The students can make an error in writing the sum and \[{{n}^{th}}\] term if they might confuse in finding the first term that is ‘a’ as the value of ‘a’ would be found only on proceeding with the question by taking the first term as an unknown variable.