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Sequence (GPREP)

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by aj5105 » Tue Jun 16, 2009 11:41 pm
Given a sequence a1, a2, a3……. a15

In the sequence shown, an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant. How many of the terms in the sequence are greater than 10?

(1) a1 = 24
(2) a8 = 10
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by bluementor » Tue Jun 16, 2009 11:52 pm
an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant

How many of the terms in the sequence are greater than 10?

Statement 1: a1 = 24

Since we dont know k, we cannot know how many terms will be greater than 10.

for eg.
-if k > 0, then all 15 terms will be greater than 10.
-if k = -20, then only the first term, a1, is greater than 10.

Insufficient.

Statement 2: a8 = 10

Notice that a8 is the middle term of the 15-term sequence. So this means there are 7 terms greater than 10, and 7 terms smaller than 10 (notice that this true regardless of what k is, as long as k is nonzero). We have sufficient info to answer this question.

Choose B.

-BM-
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by aj5105 » Wed Jun 17, 2009 12:19 am
I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?
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by shashank.mehra » Wed Jun 17, 2009 12:31 am
aj5105 wrote:I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?
1. It depends whether the common difference (in this case case K is common differnce) is positive or negative

2. As K is non-zero therefore, either the terms will be in ascending (if k is +ve) or descending (if k is -ve)
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by tohellandback » Wed Jun 17, 2009 12:37 am
aj5105 wrote:I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?
aj5105,
I agree with bluementor about the answer B
as for your questions:
1) No, a sequence is not always in an order. It depends upon the sequence's formula or function. Arithmetic or geometric progression are just examples of sequence

2) It is possible but not in this case. it is clearly mentioned that K is a non-zero constant. if K can be zero then of course all the terms are equal
The powers of two are bloody impolite!!
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by Umar82 » Wed Aug 19, 2009 2:56 pm
is there a typo in this question? how can there be 15 sequences if 2 ≤ n ≤ 15 ??
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by sumanr84 » Mon Jul 19, 2010 7:52 am
aj5105 wrote:Given a sequence a1, a2, a3��. a15

In the sequence shown, an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant. How many of the terms in the sequence are greater than 10?

(1) a1 = 24
(2) a8 = 10
Tricky question..almost got me..worth solving...
[spoiler]
takeaway : C-Trap says " In DS, Always be suspicious of answers that you can get using 2 options without using your pen i.e. too obvious looking answer." - This takeaway saved me in this question.[/spoiler]
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by kvcpk » Mon Jul 19, 2010 8:10 am
Given a sequence a1, a2, a3??. a15

In the sequence shown, an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant. How many of the terms in the sequence are greater than 10?

(1) a1 = 24

a1= 24
a2=24+k
a3=24+2k
.....
We cannot say how many are greater than 10, because, we do not know if K is positive or negative.

(2) a8 = 10
Series is linear.. Whether it needs to increase or decrease linearly.
a1,a2,..........a8,..............a14,a15
On either side of a8, there are only 7 terms.
So either a1-a7 or a9-a15 need to be less than 10.

hence pick B.

I was almost getting it wrong when I looked at why 2<=n<=15 is given. That made ot easy for me.
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by clammiestqasar » Mon Aug 02, 2010 12:18 am
we dont have 7 terms on either side of a8.Thereare 6 terms on the left of a8 and 7 to the right
2,3,4,5,6,7
9,10,11,12,13,14,15
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by kvcpk » Mon Aug 02, 2010 12:33 am
clammiestqasar wrote:we dont have 7 terms on either side of a8.Thereare 6 terms on the left of a8 and 7 to the right
2,3,4,5,6,7
9,10,11,12,13,14,15
The series is given as a1,a2,a3...
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by clammiestqasar » Mon Aug 02, 2010 2:06 am
yea..apologies for being so stupid.
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by lunarpower » Tue Aug 03, 2010 3:23 am
Umar82 wrote:is there a typo in this question? how can there be 15 sequences if 2 ≤ n ≤ 15 ??
the problem is not stating that terms only exist for 2 ≤ n ≤ 15; rather, the problem is stating that the given rule for generating terms only works when 2 ≤ n ≤ 15.
since the rule in question refers to a(n-1) and a(n), it actually implies the existence of, at the very least, terms for 1 ≤ n ≤ 15 (since plugging in n = 2 requires the existence of terms for n = 1 and n = 2).
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