Question

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

Answer 1

Given,

$a=63,d=-3,S_n=693$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$693=\frac{n}{2}\left[2\right(63)+(n-1\left)\right(-3\left)\right]$

$1386=n[126-3n+3]$

$1386=126n-3n^2+3n$

$3n^2-129n+1386=0$

$(n-22)(n-21)=0$

$\therefore n=21,22$

$a_{21}=a+(21-1)d$

$=63+(20\times (-3\left)\right)$

$=3$

$S_{21}=\frac{21}{2}[63+3]=693$

Therefore $n=21$ satisfies the condition.

$a_{22}=a+(22-1)d$

$=63+(21\times (-3\left)\right)$

$=0$

$S_{22}=S_{21}+a_{22}$

$=693+0=693$

Therefore $n=22$ also satisfies the condition.