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20 Dec 2019, 02:29
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:28) correct
21%
(01:53)
wrong
based on 242
sessions
History
Date
Time
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Not Attempted Yet
What is the sum of the multiples of 4 from 60 to 100, both inclusive?
A. 240
B. 800
C. 860
D. 880
E. 960
A. 240
B. 800
C. 860
D. 880
E. 960
20 Dec 2019, 03:52
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Answer is D: 880
Explanation:
First term =a= 60
Last term=100
Common difference=4
Last term= first term + (number of terms - 1)* common difference
100=60+ (n-1)4 ==> n=11
Sum of sequence = n/2 ( 2a + (n-1)*d) = 11/2 (2*60 + (11-1)*4) = 880
Explanation:
First term =a= 60
Last term=100
Common difference=4
Last term= first term + (number of terms - 1)* common difference
100=60+ (n-1)4 ==> n=11
Sum of sequence = n/2 ( 2a + (n-1)*d) = 11/2 (2*60 + (11-1)*4) = 880
General Discussion
20 Dec 2019, 02:36
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Bunuel
Explanation:
n=11, a = 60, d=4
Sum= n/2 {2a + (n-1)d}
=11/2*{2*60 + (11-1)4}
= 11/2 * {120+40}
= 11* 160 /2 = 880
IMO-D
