Question

72, 69, 66, \(\ldots\) Number continue in same pattern as long as they remain positive. What will be the maximum sum of terms?

• A. \(903\)
• B. \(897\)
• C. \(900\)
• D. \(882\)

Hint:

For an arithmetic progression ending at a positive last term, first determine the number of terms and then apply \[ S_n=\frac{n}{2}(a+l). \]

Answer 1

The Correct Option is C

Step 1: Spot the pattern.
The list $72, 69, 66, \ldots$ drops by $3$ each step, so it is an arithmetic progression with first term $a=72$ and common difference $d=-3$.
Step 2: Demand positive terms.
We keep adding terms only while they stay positive, so we need the general term $a_n=72-3(n-1)$ to be greater than $0$.
Step 3: Solve for the cut-off.
\[ 72-3(n-1)>0 \implies n-1<24 \implies n<25 \] So the last allowed term is the $24$th term.
Step 4: Find that last term.
\[ a_{24}=72-3(23)=72-69=3 \] The final positive term is $3$.
Step 5: Apply the sum formula.
\[ S_n=\frac{n}{2}(a+l)=\frac{24}{2}(72+3) \]
Step 6: Compute.
\[ 12\times 75 = 900 \] So the maximum sum is $900$.
\[ \boxed{900} \]