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22 Sep 2023, 01:30
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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:
55%
(hard)
Question Stats:
57% (01:39) correct
43%
(01:27)
wrong
based on 202
sessions
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Not Attempted Yet
What is the average (arithmetic mean) of a certain sequence of N consecutive integers?
(1) The sum of the terms of the series is N.
(2) The sum of the greatest and least terms is 2.
(1) The sum of the terms of the series is N.
(2) The sum of the greatest and least terms is 2.
23 Sep 2023, 10:51
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Bunuel
(1) The sum of the terms of the series is N.
This means that the number of terms = sum of the terms.
This is only possible when N=1
SUFF.
(2) The sum of the greatest and least terms is 2.
The avg. of the terms of a consecutive series can be given by :
(first term + last term )/2
SUFF.
Ans D
Hope it helps.
24 Sep 2023, 23:06
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Bunuel
Solution:
Pre Analysis:
- Sequence of N consecutive integers can be taken as an AP with common difference 1
- So, average of this sequence \(=\frac{\text{Sum of sequence}}{N}=\frac{\frac{N}{2}[2a+(N-1)\times 1]}{N}=\frac{\frac{N}{2}(2a+N-1)}{N}=\frac{2a+N-1}{2}\)
Statement 1: The sum of the terms of the series is N
- So, average of this sequence \(=\frac{\text{Sum of sequence}}{N}=\frac{N}{N}=1\)
- Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: The sum of the greatest and least terms is 2
- Sum of the greatest and least terms \(=a+a+(N-1)\times 1=a+a+N-1=2a+N-1=2\)
So, average of this sequence \(=\frac{2a+N-1}{2}=\frac{2}{2}=1\) - Thus, statement 2 alone is also sufficient
Hence the right answer is Option D
