Consider the sequence:

$72,69,66,\ldots$

The numbers continue in the same pattern as long as they remain positive. What will be the maximum possible sum of the terms of this sequence?

The sequence is:

$72,69,66,\ldots$

This is an arithmetic progression with first term:

$a=72$

Common difference:

$d=-3$

The terms continue as long as they remain positive.

Last positive term will be $3$.

Now,

$a_n=3$

Using formula:

$a_n=a+(n-1)d$

$3=72+(n-1)(-3)$

$3=72-3n+3$

$3=75-3n$

$3n=72$

$n=24$

Now, sum of first $24$ terms is:

$S_n=\frac{n}{2}(a+l)$

$S_{24}=\frac{24}{2}(72+3)$

$S_{24}=12\times 75$

$S_{24}=900$