Consider the sequence: 72, 69, 66, ... 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?
Step-by-step Solution:
The given sequence is an Arithmetic Progression (AP) with the first term $a = 72$ and common difference $d = -3$. For the terms to remain positive, the $n$-th term must be greater than 0: $$ T_n = a + (n-1)d > 0 $$ $$ 72 - 3(n-1) > 0 $$ $$ 3(n-1) < 72 $$ $$ n-1 < 24 \implies n < 25 $$ The maximum number of positive terms is 24. The sum of the first 24 terms is: $$ S_{24} = \frac{n}{2} [2a + (n-1)d] $$ $$ S_{24} = \frac{24}{2} [2(72) + (23)(-3)] $$ $$ S_{24} = 12 [144 - 69] = 12 \times 75 = 900 $$ Hence, the maximum sum is 900.