Question

How many terms of the AP 21, 18, 15, ... must be added to get the sum 0?

Answer 1

The given AP is 21, 18, 15, ... .

Here, a = 21 and d = 18 − 21 = −3

Let the required number of terms be n. Then,

$$\begin{gathered} S_n=0 \\ \Rightarrow \frac{n}{2}\left[2\times 21+\left(n-1\right)\times \left(-3\right)\right]=0\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow \frac{n}{2}\left(42-3n+3\right)=0 \\ \Rightarrow n\left(45-3n\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow n=0\mathrm{or}45-3n=0 \\ \Rightarrow n=0\mathrm{or}n=15 \end{gathered}$$

∴ n = 15 (Number of terms cannot be zero)

Hence, the required number of terms is 15.