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:
$S_n=\frac{n}{2}(a+l)$
$S_{24}=\frac{24}{2}(72+3)$
$S_{24}=12\times 75$
$S_{24}=900$
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