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tough sequence problem

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tough sequence problem

by vkb16 » Tue Sep 08, 2009 1:11 am
The sequence A1, A2, A3... An of n such integers is such that Ak = k if k is odd, and Ak = -A(k-1) if K is even. Is the sum of the terms in the sequence positive?

1. n is odd
2. An if positive

note- -A(k-1)<< here, k-1 is the term number, Not multiplied

Whats the best way to go about in trying to solve these types of sequence sums?

Thanks in advance!
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by sanjib » Tue Sep 08, 2009 2:56 am
Unless we know what is A1 we dont know the answer.But we do get A1 = 1 from St.1
and we do get information of An from St.2
so IMO C
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Re: tough sequence problem

by Ian Stewart » Tue Sep 08, 2009 4:01 am
vkb16 wrote:The sequence A1, A2, A3... An of n such integers is such that Ak = k if k is odd, and Ak = -A(k-1) if K is even. Is the sum of the terms in the sequence positive?

1. n is odd
2. An if positive

note- -A(k-1)<< here, k-1 is the term number, Not multiplied

Whats the best way to go about in trying to solve these types of sequence sums?

Thanks in advance!
It's almost always a good idea to write down the first few terms of the sequence - often a pattern will emerge. We know that A_k = k when k is odd, so A_1 = 1, A_3 = 3, A_5 = 5, etc. When k is even, A_k = -A_(k-1), so

A_2 = -A_1 = -1
A_4 = -A_3 = -3
A_6 = -A_5 = -5, etc

and the sequence is: 1, -1, 3, -3, 5, -5, 7, -7, ...

Now you can see that if you add an even number of terms, the sum will always be zero, whereas if you add an odd number of terms, the sum will be positive. So Statement 1 is sufficient alone. From Statement 2, if A_n is positive, then n must be odd, so Statement 2 gives the same information as Statement 1, and is also sufficient. The answer is D.
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