Mathematics — Standard (041) — CBSE Class 10 board question

Given: $a_n = 3n - 2$

First term: $a_1 = 3(1) - 2 = 1$

Last term (28th): $a_{28} = 3(28) - 2 = 84 - 2 = 82$

Using $S_n = \dfrac{n}{2}(a + l)$:

$$S_{28} = \frac{28}{2}(1 + 82) = 14 \times 83 = \mathbf{1162}$$

Source: Chapter 5, Summary Point 5

• Find first term by putting $n = 1$, and last (28th) term by putting $n = 28$.
• Use the formula $S = \frac{n}{2}(a + l)$ since both first and last terms are known — this is quicker than using $S = \frac{n}{2}[2a + (n-1)d]$.
• Both approaches are acceptable; examiners award marks for correct formula, substitution, and final answer.