Find the sum of first 20 terms of an AP in which $d = 5$ and $a_{20} = 135$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding rag
Model Answer
Given: $d = 5$, $a_{20} = 135$, $n = 20$
Using $a_n = a + (n-1)d$:
$135 = a + 19 \times 5$
$a = 135 - 95 = 40$
Now, $S_{20} = \dfrac{n}{2}(a + l) = \dfrac{20}{2}(40 + 135) = 10 \times 175 = \mathbf{1750}$
Source: Chapter 5, Exercise 5.3
---
Explanation
- First find the first term $a$ using $a_n = a + (n-1)d$.
- Then use $S = \frac{n}{2}(a + l)$ where $l = a_{20} = 135$ (the last term here), which is simpler than using $S = \frac{n}{2}[2a + (n-1)d]$.
- Both formulas give the same answer; examiners award 1 mark for finding $a$ and 1 mark for the correct sum.