Question

Find the number of terms of the AP 18, 15, 12, ... so that their sum is 45. Explain the double answer.

Answer 1

Here, a = 18 and d = (15- 18) = -3

Let the required number of terms be n.
Then, S_n = 45
⇒ $\frac{n}{2}[2a+\left(n-1\right)d]$ = 45

$$\begin{gathered} \Rightarrow \left(\frac{n}{2}\right)\times \left[2\times 18+\left(n-1\right)\times \left(-3\right)\right]=45 \\ \Rightarrow \left(\frac{n}{2}\right)\times \left[39-3n\right]=45 \\ \Rightarrow 3n^2-39n+90=0 \\ \Rightarrow n^2-13n+30=0 \\ \Rightarrow n^2-10n-3n+30=0 \\ \Rightarrow n(n-10)-3(n-10)=0 \\ \Rightarrow (n-10)(n-3)=0 \\ \Rightarrow n=3orn=10 \end{gathered}$$

∴ Sum of the first 3 terms = sum of the first 10 terms = 45
This means that the sum of all terms from 4^th to 10^th is zero.