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11 May 2020, 23:34
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
60% (02:43) correct
40%
(01:41)
wrong
based on 10
sessions
History
Date
Time
Result
Not Attempted Yet
How many terms at the maximum, of the progression 2,5,8,........can be considered, if the sum of the terms is to be less than 3000?
A.60
B.44
C.68
D.70
E.67
A.60
B.44
C.68
D.70
E.67
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11 May 2020, 23:55
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MBADream786
CONCEPT:Sum of an Arithmetic Progression = (n/2)*(First-Term + Last-Term) = (n/2)*[2a+(n-1)*d]
Here, First term, a = 2
Common Difference, d = 5-2 = 8-5 = 3
i.e. Sum of an Arithmetic Progression = (n/2)*[2*2+(n-1)*3] < 3000 (Given)
i.e. Sum of an Arithmetic Progression = (n/2)*[4+(3n-3)] < 3000
i.e. Sum of an Arithmetic Progression = (n/2)*(3n+1) < 3000
i.e. (n)*(3n+1) < 6000
Let,s Check Options:
A.60, i.e. n = 60, (n)*(3n+1) = 60*181 > 6000 OUT and all options other than B are OUT too
B.44, i.e. n = 4, (n)*(3n+1) = 44*133 = 5852 < 6000 CORRECT
C.68
D.70
E.67
ANswer: Option B
Yogananda
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12 May 2020, 00:05
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Thanks a lot for the quick reply.
