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DS - sequence

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DS - sequence

by Xbond » Sat Aug 29, 2009 8:03 am
Hi there,

Could you explain in the simplest this concept and how to resolve it


In a sequence, a1, a2, ... an, a1 = 64, a2 = 66, a3 = 67. For all n > 3, an = a(n-3) + 8.
Which of the following is a term of the sequence?

a)762
b)765
c)801
d)1006
e)1287
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by real2008 » Sat Aug 29, 2009 9:15 am
is it A
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by pd0483 » Sat Aug 29, 2009 9:20 am
The answer is A i.e. 762

For n>3, an=a(n-3)+8.
Therefore, [an-a(n-3)] should be divisible by 8.
762-66=696 which is divisible by 8.
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by rebelsid » Sat Aug 29, 2009 10:44 pm
y did you take 762-66?
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by pd0483 » Sun Aug 30, 2009 6:37 am
I did took 66 by trial and error, but this method is time consuming. If anyone has a better method, they can share it with us.
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by Xbond » Sun Aug 30, 2009 7:57 am
oa answser is A
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by bharathh » Sun Aug 30, 2009 10:29 am
The answer is A.

Although I did manage to evaluate this question eventually there is no way I could have done in during the actual test if I had got it. What level of difficulty is associated with this question?

Also how do you identify a question like this?

Here's my solution

a1 = 64 = 8.8
a2 = 66 = 8.8 + 2
a3 = 67 = 8.8 + 3
a4 = 72 = 8.8 + 8
a5 = 74 = 8.8 + 8 + 2
a6 = 75 = 8.8 + 8 + 3
a7 = 8.8 + 8.2
a8 = 8.8 + 8.2 + 2
a9 = 8.8 + 8.2 + 3
and so on

Any number that gives us a remainder of 0,2 or 3 when divided by 8 will be in this series.

Of the answer choices,

762/8 r = 2 .. In sequence
765/8 r = 5 .. Not in the sequence
801/8 r = 1 .. Not in sequence
1006/8 r = 4 .. Not in sequence
1287/8 r = 7 .. Not in sequence.

So answer is A
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by pathaniaus » Sun Aug 30, 2009 11:31 am
I seriously don't understand this question.. can someone please explain it in a very simple way?

Thank you!
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