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

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

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• 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.