Given \(a_3 = 12\) and \(a_{50}= 106\).
The formula for the nth term of an arithmetic sequence is \(a_n = a + (n − 1) d\).
Using this formula for \(a_3\), we get \(a_3 = a + (3 − 1) d\), which simplifies to \(12 = a + 2d\) (Equation i).
Similarly, for \(a_{50}\), we have \(a_{50} = a + (50 − 1)d\), which simplifies to \(106 = a + 49d\) (Equation ii).
Subtracting Equation (i) from Equation (ii) yields \(94 = 47d\).
Solving for \(d\), we find \(d = 2\).
Substituting \(d = 2\) back into Equation (i), we get \(12 = a + 2 \times 2\).
Simplifying, \(a = 12 − 4\), so \(a = 8\).
To find the 29th term, we use the formula: \(a_{29} = a + (29 − 1)d\).
Substituting the values of \(a\) and \(d\), we get \(a_{29} = 8 + 28\times 2\).
This calculates to \(a_{29} = 8 + 56 = 64\).
Therefore, the 29th term is 64.