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# Given a sequence: a_1, a_2, a_3, ... a_{14}, a_{15} In the

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Manager
Joined: 22 Jul 2006
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Given a sequence: a_1, a_2, a_3, ... a_{14}, a_{15} In the [#permalink]

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04 Nov 2006, 04:38
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Question Stats:

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Given a sequence: $$a_1, a_2, a_3, ... a_{14}, a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-sequence-shown-a-n-a-n-1-k-where-2-n-15-and-k-126119.html
[Reveal] Spoiler: OA
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04 Nov 2006, 06:04
A8 IS THE MIDDLE TERM
Manager
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04 Nov 2006, 14:52
Again good point Joey.,

The question should read k a non-zero constant...!
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05 Nov 2006, 15:37
yezz wrote:
A8 IS THE MIDDLE TERM

Yezz, how did you come to a conclusion that A8 is the middle term ?
The sequence has 14 terms , how is A8 the middle term ? What am I missing here ?
Senior Manager
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05 Nov 2006, 22:45
seq has 15 terms..note that a1 can also exist.

but my point is that when k=0,there r no terms greater than 10 becausethen all r 10.

but when k>0 or <0...u've exactly 7 terms which r greater than 10.

so unless,it is mentioned that k is non zero,i think ans should be E
and for k not equal to zero,ans should be B

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06 Nov 2006, 01:45
seq has 15 terms..note that a1 can also exist.

but my point is that when k=0,there r no terms greater than 10 becausethen all r 10.

but when k>0 or <0...u've exactly 7 terms which r greater than 10.

so unless,it is mentioned that k is non zero,i think ans should be E
and for k not equal to zero,ans should be B

no terms are greater than 10 is an answer = 0
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06 Nov 2006, 01:53
if K is non zero, then B.

if it does not say so, then C.
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16 Jun 2009, 05:49
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Answer is C. I dont understand how it could be B if K must not be 0?
The set has 14 numbers. "A8" is the 7th of those 14 numbers. In any with an even amount of numbers there is no middle number and you are therefore not able to decide if there are 6 or 7 numbers >10.

It does not matter if they give you A7 and 1<= n <=14, A8 and 2<= n <=15 or A86 and 80<= n <=93. As long as the set consists of an even amount of numbers there is no middle number.
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18 Jun 2009, 10:18
The Question is mis transcribed. Original Q says K!=0
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22 Aug 2009, 22:30
An = An-1 + K for K! = 0 would be arithmetic expression with K as the difference.
Hence it can be concluded from second expression(A8=10) that in a decreasing sequence number ,
7th term is the closest term to the value of 10.
7 terms have value more than 10.

more detailed analysis (though Statement 2 is enough for conclusion)
putting in A1 = 10 and A8 = 24,

A8 = A1 + (8-1)K -- ( n term of the sequence = A1 + (n-1)K )
10 = A1 + 7K
10 = 24 + 7K
K = - 2

The value of 7th term is A7 = A1 + 6k = 24 - 12 = 12 > 10 .
----------------------

Statement 1 is not sufficient enough /
Statement 2 is sufficient
Statement 1 + 2 = confirm the values further
Manager
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23 Aug 2009, 00:10
since K value is not mentioned in the question,
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Kudos me if my reply helps!

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23 Aug 2009, 01:57
IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.
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23 Aug 2009, 04:41
rohansherry wrote:
IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on...
Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.
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23 Aug 2009, 09:04
Assumption: K is a constant according the defined sequence nth term.
Since A1 = 24 and A8 = 10, K is negative in a decreasing sequence. 7 terms on the left side are >10

If K is not a constant but some function of the terms like K = n^((-1)^n) which will give fractions in case of odd terms and whole integers in case of even.
or K = (-1)^n in which case the sequence can have positive and negative numbers based on the value of n.
This will lead the solution being E as both statements are insufficient.
But I think that assumption made about K being constant is correct according to the definition of a sequence.

Economist wrote:
rohansherry wrote:
IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on...
Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.
Manager
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23 Aug 2009, 09:46
Economist wrote:
rohansherry wrote:
IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on...
Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.

Why don't you consider k=0? If k=0, then a1=a2=...=a8=a9=...=a15=10...so no inetger is greater than 10...so the answer could be:
1) zero if k= 0
2) 7 if K is either positive or negative

Combining statements 1 and 2, it can be concluded that k is negative. Sufficient. There are 7 numbers greater than 10.
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23 Aug 2009, 09:55
icandy wrote:
The Question is mis transcribed. Original Q says K!=0

K! = 0 is not correct. Only K = 0 is possible. If K = 0, K! = 1.
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14 Apr 2010, 11:55
Can someone please explain how there are 15 terms in this set? I can only see 14 if n is truly >= 2 and <=15.
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14 Apr 2010, 13:52
shahany wrote:
Can someone please explain how there are 15 terms in this set? I can only see 14 if n is truly >= 2 and <=15.

I edited the question above, there was one part missing.

We have a sequence of fifteen terms (actually this sequence is arithmetic progression). As k is nonzero, all elements would be different and the median would be the eighth term, $$a_8$$. This means that 7 terms will be less than $$a_8$$ and 7 terms will be more than $$a_8$$. Note here that it doesn't matter whether k is positive or negative:

if k is positive, we'll get an ascending sequence and the terms from from $$a_1$$ to $$a_7$$ will be less than $$a_8$$ and terms from $$a_9$$ to $$a_{15}$$ will be more than $$a_8$$;

if k is negative, we'll get an descending sequence and the terms from from $$a_1$$ to $$a_7$$ will be more than $$a_8$$ and terms from $$a_9$$ to $$a_{15}$$ will be less than $$a_8$$.

Statement (1) is giving the value of $$a_1$$, but since we don't know k, we can not say how many terms are more than 10: it can vary from 1 (only $$a_1=24>10$$, if k<=-14) to 15 (if k is positive for instance).

Statement (2) is saying that $$a_8=10$$. As we discussed above, $$a_8$$ is median value and for any value of k, 7 terms will be more than $$a_8=10$$ and 7 terms will be less than $$a_8=10$$. Hence this statement is sufficient.

Hope it helps.
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18 Sep 2010, 09:31
Nice question !!

I would have definitely marked C in the exam.
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Re: Sequence OG, DS   [#permalink] 18 Sep 2010, 09:31
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# Given a sequence: a_1, a_2, a_3, ... a_{14}, a_{15} In the

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