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Sequence

by dhairya275 » Sun Sep 09, 2012 8:19 am
A man writes a special sequence of 11 numbers in ascending order such that the difference between any two consecutive numbers of the series is same.What is the median of these numbers given that all the numbers are positive integers?
(1) The sum total of all these 11 numbers is 121.
(2) The ratio of 3rd number to the 8th number from left side is 1/3

Please Help !!
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Source: — Data Sufficiency |
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by KapilKapoor » Sun Sep 09, 2012 9:19 am
If the difference between any two consecutive numbers is same so it is kind of Arithmetic progression.
The median will be 6th term and the sum of the 11 numbers will be 11*Avg of the 11 numbers.

St 1: 11*Avg=121
Avg = Median = T6 = sixth terms = 11
St 2: a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11
a3/a8=1/3
(a1+2d)/(a1+7d)=1/3
3(a1)+6d=a1+7d
2(a1)=d
T6= median = a1+5d= 11a1
Insufficient.

Regards,
Kapil
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by Javoni » Sun Sep 09, 2012 11:45 am
Let the the first number of sequence be A, the next would be A+X, A+2X, A+3X and so forth where the 11th member would be A+10X.

(1) We get that Sum of total is 121, hence 11A+55X=121, thus A+5X = 11, which is the median of the sequence above, i.e. the 6th member of the sequence. Thus Number 1 SUFFICIENT.

(2) It says 3d member/8th member = 1/3, or or to be precise (A+2X)/(A+7X) = 1/3, means 2A=X, nothing much, thus INSUFFICIENT.

Therefore, answer I guess would be (A)

Please, please correct me if I went awry, thanks!

dhairya275 wrote:A man writes a special sequence of 11 numbers in ascending order such that the difference between any two consecutive numbers of the series is same.What is the median of these numbers given that all the numbers are positive integers?
(1) The sum total of all these 11 numbers is 121.
(2) The ratio of 3rd number to the 8th number from left side is 1/3

Please Help !!
Life begins at the End of your Comfort Zone...
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Junior | Next Rank: 30 Posts
Posts: 24
Joined: Sat Jul 09, 2011 1:45 pm
Location: Tajikistan, Dushanbe
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by Javoni » Sun Sep 09, 2012 11:46 am
Let the the first number of sequence be A, the next would be A+X, A+2X, A+3X and so forth where the 11th member would be A+10X.

(1) We get that Sum of total is 121, hence 11A+55X=121, thus A+5X = 11, which is the median of the sequence above, i.e. the 6th member of the sequence. Thus Number 1 SUFFICIENT.

(2) It says 3d member/8th member = 1/3, or or to be precise (A+2X)/(A+7X) = 1/3, means 2A=X, nothing much, thus INSUFFICIENT.

Therefore, answer I guess would be (A)

Please, please correct me if I went awry, thanks!

dhairya275 wrote:A man writes a special sequence of 11 numbers in ascending order such that the difference between any two consecutive numbers of the series is same.What is the median of these numbers given that all the numbers are positive integers?
(1) The sum total of all these 11 numbers is 121.
(2) The ratio of 3rd number to the 8th number from left side is 1/3

Please Help !!
Life begins at the End of your Comfort Zone...
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