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13 Mar 2020, 15:58
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D
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Difficulty:
45%
(medium)
Question Stats:
86% (02:04) correct
14%
(02:47)
wrong
based on 14
sessions
History
Date
Time
Result
Not Attempted Yet
What is the sum of all 3 digit positive integers that leave a remainder of 2 when divide by 3?
A. 163900
B. 163751
C. 164700
D. 164850
E. 165000
A. 163900
B. 163751
C. 164700
D. 164850
E. 165000
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firas92
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13 Mar 2020, 18:45
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First term = 101
Last term = 998
Common difference = 3
Number of terms = \(\frac{(998-101)}{3}+1=300\)
Sum of terms = \(\frac{300}{2}(101+998)=164850\)
Answer is (D)
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Last term = 998
Common difference = 3
Number of terms = \(\frac{(998-101)}{3}+1=300\)
Sum of terms = \(\frac{300}{2}(101+998)=164850\)
Answer is (D)
Posted from my mobile device
globaldesi
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13 Mar 2020, 20:59
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total number of numbers are
101, 104 ....998
hence
998-101 = 299+1= 300 (since both are included)
\(sum of a AP = \frac{no of terms}{ 2} *( first term+last term)\)
= 300/2 (101+998)
=164850
D
101, 104 ....998
hence
998-101 = 299+1= 300 (since both are included)
\(sum of a AP = \frac{no of terms}{ 2} *( first term+last term)\)
= 300/2 (101+998)
=164850
D
