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

Given: $a_n = 5 - 2n$

First term: $a_1 = 5 - 2(1) = 3$

Common difference: $d = a_2 - a_1 = [5-2(2)] - 3 = 1 - 3 = -2$

Sum of first 20 terms:

$$S_{20} = \frac{n}{2}[2a + (n-1)d] = \frac{20}{2}[2(3) + (20-1)(-2)]$$

$$= 10[6 + 19 \times (-2)] = 10[6 - 38] = 10 \times (-32) = \mathbf{-320}$$

Source: Chapter 5, Exercise 5.3

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• First find $a$ and $d$ from the given $a_n$ formula before applying the sum formula.
• The examiner expects you to show the values of $a$, $d$, and then substitute into $S_n = \frac{n}{2}[2a+(n-1)d]$.
• A common mistake is forgetting the negative sign in $d = -2$, which changes the answer entirely.