How many terms of the A.P: 24,21,18,… must be taken so that their sum is 78

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

So, $78=\frac{n}{2}\left[48+\left(n-1\left(-3\right)\right)\right]=\frac{n}{2}\left[51-3n\right]$

Or $3n^2-51n+156=0$

Or $n^2-17n+52=0$

Or $\left(n-4\right)\left(n-13\right)=0$

Or $n=4\text{or}13$

Both values of $n$  are asmissible. So, the number of terms is either 4 or 13.