# How many terms of the AP 63, 60, 57, 54, ...

Question:

How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.

Solution:

The given AP is 63, 60, 57, 54, ... .

Here, a = 63 and d = 60 − 63 = −3

Let the required number of terms be n. Then,

$S_{n}=693$

$\Rightarrow \frac{n}{2}[2 \times 63+(n-1) \times(-3)]=693 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$\Rightarrow \frac{n}{2}(126-3 n+3)=693$

$\Rightarrow n(129-3 n)=1386$

$\Rightarrow 3 n^{2}-129 n+1386=0$

$\Rightarrow 3 n^{2}-66 n-63 n+1386=0$

$\Rightarrow 3 n(n-22)-63(n-22)=0$

$\Rightarrow(n-22)(3 n-63)=0$

$\Rightarrow n-22=0$ or $3 n-63=0$

$\Rightarrow n=22$ or $n=21$

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.

$a_{22}=63+(22-1) \times(-3)=63-63=0$

Hence, the required number of terms is 21 or 22.