How many terms of the AP $24, 21, 18, \ldots$ must be taken so that their sum is 78?

Use the sum formula $S_n=\frac{n}{2}[2a+(n-1)d]$ with $a=24$ and $d=-3$. $$78=\frac{n}{2}[48+(n-1)(-3)] = \frac{n}{2}(51-3n).$$ So $156=n(51-3n)$, giving $3n^2-51n+156=0$. Dividing by 3, $n^2-17n+52=0=(n-4)(n-13)$. Hence $n=4$ or $n=13$, so both are possible.