The 10th term of an AP with first term $a = 5$ and common difference $d = 3$ is

1. AP formula: $a_n = a + (n-1)d$. 2. Substitute $a = 5$, $d = 3$, $n = 10$. 3. $a_{10} = 5 + (10-1)(3) = 5 + 9 \cdot 3 = 5 + 27 = 32$. 4. Quick verification by counting: starts at 5, add 3 nine times → 5+3=8, 11, 14, 17, 20, 23, 26, 29, 32. ✓ 5. Option A uses $n$ instead of $(n-1)$. Option C overshoots by 1. Option D adds 30 instead of 27. _Source: NCERT Class 11 Mathematics, Ch 8, §8.2 + Examples, p. 3–4._