Question

Find the number of terms of the AP 64, 60, 56, ... so that their sum is 544. Explain the double answer.

Answer 1

Here, a = 64 and d = (60 - 64) = -4

Let the required number of terms be n.
Then, S_n = 544
⇒ $\frac{n}{2}[2a+\left(n-1\right)d]$ = 544

$$\begin{gathered} \Rightarrow \left(\frac{n}{2}\right)\times \left[2\times 64+\left(n-1\right)\times \left(-4\right)\right]=544 \\ \Rightarrow \left(\frac{n}{2}\right)\times \left[132-4n\right]=544 \\ \Rightarrow 2n^2-66n+544=0 \\ \Rightarrow n^2-33n+272=0 \\ \Rightarrow n^2-17n-16n+272=0 \\ \Rightarrow n(n-17)-16(n-17)=0 \\ \Rightarrow (n-17)(n-16)=0 \\ \Rightarrow n=17orn=16 \end{gathered}$$

∴ Sum of the first 16 terms = sum of the first 17 terms = 544
This means that the term T₁₇ is 0.