Step 1
Given:
Second term \((a_{2}) =\ -4\)
Third term \((a_{3}) = 12\)
Step 2
Used concept
\(T_{n} = a_{1}\ +\ (n\ -\ 1)d\)
Where \(T_{n}\ \rightarrow\ n^{(th)}\ \text{term}\)
\(a_{1}\ \rightarrow\ 1^{(st)}\ \text{term}\)
\(d\ \rightarrow\ \text{difference}\ = (a_{2}\ -\ a_{1}) = (a_{3}\ -\ a_{2})\) Step 3 Apply the above concept it gives \(d = a_{3}\ -\ a_{2}\)
\(d = 12\ -\ (-4)\)
\(d = 12\ +\ 4 = 16\) now, \(a_{1} = a_{2}\ -\ d\)
\(a_{1} =\ -4\ -\ 16 =\ -20\) Step 4 The \(37^{(th)}\) term of the given arithmetic sequence will be \(T_{37} =\ -20\ +\ (37\ -\ 1)\ \times\ 16\)
\(=\ -20\ +\ 36\ \times\ 16\)
\(=\ -20\ +\ 576\)
\(=556\) (answer)
\(a_{1}\ \rightarrow\ 1^{(st)}\ \text{term}\)
\(d\ \rightarrow\ \text{difference}\ = (a_{2}\ -\ a_{1}) = (a_{3}\ -\ a_{2})\) Step 3 Apply the above concept it gives \(d = a_{3}\ -\ a_{2}\)
\(d = 12\ -\ (-4)\)
\(d = 12\ +\ 4 = 16\) now, \(a_{1} = a_{2}\ -\ d\)
\(a_{1} =\ -4\ -\ 16 =\ -20\) Step 4 The \(37^{(th)}\) term of the given arithmetic sequence will be \(T_{37} =\ -20\ +\ (37\ -\ 1)\ \times\ 16\)
\(=\ -20\ +\ 36\ \times\ 16\)
\(=\ -20\ +\ 576\)
\(=556\) (answer)
