Given \(a_{11} = 38\) and \( a_{16} = 73\).
The formula for the nth term of a sequence is \(a_n = a + (n − 1) d\).
For the 11th term: \(a_{11} = a + (11 − 1) d \), which simplifies to \(38 = a + 10d \) ……..(1).
For the 16th term: \(a_{16} = a + (16 − 1) d \), which simplifies to \(73 = a + 15d \) …….(2).
Subtracting equation (1) from equation (2) yields \(35 = 5d\), so \(d = 7\).
Substituting \(d = 7\) into equation (1): \(38 = a + 10(7)\).
This gives \(38 = a + 70\), so \(a = 38 − 70 = −32\).
To find the 31st term: \(a_{31} = a + (31 − 1) d\).
Substituting the values of \(a\) and \(d\): \(a_{31}= − 32 + 30 (7) \).
Calculating: \(a_{31}= − 32 + 210\).
Therefore, \(a_{31}= 178\).
The \(31^{st}\) term is \(178\).