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GATE CS 2013 · General Aptitude

Question 58

MCQPrinted as GA Q.581 markWrong answer: −1/3Easy (estimate)

Question

What will be the maximum sum of 44,42,40,…44, 42, 40, \dots?

Choose one option
Official answer and solution
Official answer
C
From
Official GATE 2013 answer key

Solution

Find the maximum sum of the decreasing arithmetic progression by summing all non-negative terms.

The given series is an arithmetic progression (AP):

44,42,40,…44, 42, 40, \dots
  • First term a=44a = 44.
  • Common difference d=42−44=−2d = 42 - 44 = -2.

Because the terms decrease by 2 at each step, adding positive terms increases the sum, adding zero leaves the sum unchanged, and adding negative terms decreases the sum. To maximize the sum, we sum all non-negative terms:

Tn=a+(n−1)d=44−2(n−1)=46−2n≥0T_n = a + (n - 1)d = 44 - 2(n - 1) = 46 - 2n \ge 0
2n≤46  ⟹  n≤232n \le 46 \implies n \le 23
  • The 22nd term is T22=46−44=2T_{22} = 46 - 44 = 2.
  • The 23rd term is T23=46−46=0T_{23} = 46 - 46 = 0.

The maximum sum is obtained by summing either the first 22 or the first 23 terms:

S22=222(a+T22)=11×(44+2)=11×46=506S_{22} = \dfrac{22}{2}(a + T_{22}) = 11 \times (44 + 2) = 11 \times 46 = 506
S23=232(a+T23)=232×44=23×22=506S_{23} = \dfrac{23}{2}(a + T_{23}) = \dfrac{23}{2} \times 44 = 23 \times 22 = 506

Thus, the maximum sum is 506, which corresponds to option (C).

Final answer

C

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2013 Computer Science and Information Technology (CS)
Question
58 of 65, printed as GA Q.58
Section
General Aptitude

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