Question

Find the number of terms of the AP 63, 60, 57, ... so that their sum is 693. Explain the double answer.

Answer 1

Here, a = 63 and d = (60 - 63) = -3

Let the required number of terms be n.
Then, S_n = 693
⇒ $\frac{n}{2}[2a+\left(n-1\right)d]$ = 693

$$\begin{gathered} \Rightarrow \left(\frac{n}{2}\right)\times \left[2\times 63+\left(n-1\right)\times \left(-3\right)\right]=693 \\ \Rightarrow \left(\frac{n}{2}\right)\times \left[129-3n\right]=693 \\ \Rightarrow 3n^2-129n+1386=0 \\ \Rightarrow n^2-43n+462=0 \\ \Rightarrow n^2-21n-22n+462=0 \\ \Rightarrow n(n-21)-22(n-21)=0 \\ \Rightarrow (n-21)(n-22)=0 \\ \Rightarrow n=21orn=22 \end{gathered}$$

∴ Sum of the first 21 terms = sum of the first 22 terms = 693
This means that the term T₂₂ is 0.