Question

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?

• A. 900
• B. 897
• C. 882
• D. 903

Hint:

When an AP ends, the last term is usually the smallest positive value in the sequence.

Answer 1

The Correct Option is A

Step 1: Recognise the pattern.
The terms 72, 69, 66, ... drop by 3 each time, so this is an arithmetic progression with first term $a = 72$ and common difference $d = -3$.
Step 2: Decide where to stop.
We keep terms only while they stay positive. The terms are $72, 69, \dots, 6, 3$. The next would be $0$, which is not positive, so the last term we keep is $3$.
Step 3: Count how many terms there are.
Use $a_n = a + (n-1)d$. Setting $a_n = 3$: $3 = 72 + (n-1)(-3)$, so $(n-1)(-3) = -69$, giving $n - 1 = 23$ and $n = 24$. There are 24 terms.
Step 4: Apply the sum formula.
$S_n = \dfrac{n}{2}(a + a_n)$ uses just the first and last terms.
Step 5: Substitute the values.
$S_{24} = \dfrac{24}{2}(72 + 3) = 12 \times 75$.
Step 6: Compute the total.
$12 \times 75 = 900$. So the largest possible sum while terms remain positive is 900, matching option 1.
\[ \boxed{900} \]