1. **Problem statement:** Given the sequence 5, 11, 17, 23, 29, find: (a) the expression for the nth term, (b) the sum of the first 10 terms, (c) which term equals 119. 2. **Identify the sequence type:** The sequence increases by 6 each time (11-5=6, 17-11=6), so it is an arithmetic sequence. 3. **Formula for nth term of an arithmetic sequence:** $$a_n = a_1 + (n-1)d$$ where $a_1$ is the first term and $d$ is the common difference. 4. **Find nth term expression:** Given $a_1=5$ and $d=6$, $$a_n = 5 + (n-1)6 = 5 + 6n - 6 = 6n - 1$$ 5. **Sum of first n terms formula:** $$S_n = \frac{n}{2}(a_1 + a_n)$$ 6. **Sum of first 10 terms:** Calculate $a_{10}$: $$a_{10} = 6(10) - 1 = 60 - 1 = 59$$ Then, $$S_{10} = \frac{10}{2}(5 + 59) = 5 \times 64 = 320$$ 7. **Find which term equals 119:** Set $a_n = 119$: $$6n - 1 = 119$$ Add 1 to both sides: $$6n = 120$$ Divide both sides by 6: $$n = \frac{\cancel{6}n}{\cancel{6}} = \frac{120}{6} = 20$$ **Final answers:** (a) $a_n = 6n - 1$ (b) $S_{10} = 320$ (c) The 20th term is 119.