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

Concept: The given sequence is an Arithmetic Progression (A.P.) because the difference between consecutive terms is constant. \[ 69-72=-3 \] \[ 66-69=-3 \] Hence, \[ a=72,\qquad d=-3 \] The sequence continues until the terms remain positive.

Step 1: Find the last positive term. General term: \[ a_n=a+(n-1)d \] \[ a_n=72-3(n-1) \] Since terms must remain positive, \[ 72-3(n-1)>0 \] \[ 24-(n-1)>0 \] \[ n<25 \] Thus, \[ n=24 \] The last positive term is \[ a_{24}=72-3(23)=3 \] Hence sequence is \[ 72,69,66,\ldots,3 \]

Step 2: Find the sum of all terms. Using A.P. sum formula, \[ S_n=\frac{n}{2}(a+l) \] where \[ n=24,\quad a=72,\quad l=3 \] Therefore, \[ S_{24} = \frac{24}{2}(72+3) \] \[ = 12\times75 \] \[ =900 \] Thus, \[ \boxed{900} \] Therefore, \[ \boxed{\text{Answer = (C)}} \]