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| OG - PS - #247 |
| #247 |
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Jeff
Guest
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Hi -
Did you forget to post the original sequence? /Jeff |
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Stacey Koprince
MGMAT STAFF
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Yep! mww7786, can you please post the full question text when you get a chance? Thanks!
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| OG #247 |
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mww7786
Guest
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OG #247
Sequece: p,r,s,t,u An arithmetic sequence is a sequence in which each term after the first is equal tot hte sum of the preceding term and a constant. If the list numbers shown above is an arithmetic sequence which of the following must also be an arithmetic sequence I. 2p, 2r, 2s, 2t, 2u II. p-, r-3, s-3, t-3, u-3 III. P^2, r^2, s^2, t^2, u^2 A I B II C III D I and II e. II and III |
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Jadran Lee
MGMAT STAFF
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Given information:
a) p,r,s,t,u is an arithmetic sequence b) An arithmetic sequence is a sequence (i.e. a list of numbers) in which each term after the first is equal to the sum of the preceding term and a constant We begin by translating this given information into algebra. We can say that there is a constant c such that p+c=r, and r+c=s, and s+c=t, and t+c=u. We can rewrite this more concisely by saying: there is a constant c such that c=r-p=s-r=t-s=u-t Now let's look at the question. It asks which of the following must be an arithmetic sequence: I. 2p, 2r, 2s, 2t, 2u II. p-3, r-3, s-3, t-3, u-3 III. p^2, r^2, s^2, t^2, u^2 We have to use our definition of an arithmetic sequence once again. We can rephrase the question, as it relates to (I), to ask (I) Does there have to be some constant d (I'm calling it "d" because different series need not have the same constant) such that d=2r-2p=2s-2r=2t-2s=2u-2t ? The OG explanation does a good job of showing why the answer to this question is YES, and that d just equals 2c. Rephrasing the question as it relates to (II) gives us (II) Does there have to be some constant e such that e=(r-3)-(p-3)=(s-3)-(r-3)=(t-3)-(s-3)=(u-3)-(t-3) ? Clearly, (r-3)-(p-3) = r-p = c, and so on for the other terms in this sequence, so the answer is again YES. Rephrasing the question as it relates to (III) gives us: (II) Does there have to be some constant f such that f=(r^2)-(p^2)=(s^2)-(r^2)=(t^2)-(s^2)=(u^2)-(t^2) ? Here the answer is NO. To see why, let's look at (r^2)-(p^2). We know that r=p+c, so r^2=p^2 + 2pc + c^2. That means (r^2)-(p^2) = 2pc + c^2. Let's compare this to (s^2)-(r^2). We know that s=r+c, so s^2=r^2 + 2rc + c^2. That means (s^2)-(r^2) = 2rc + c^2. Our rephrased question asked whether it MUST be true that (r^2)-(p^2)=(s^2)-(r^2). This is asking whether it MUST be true that 2pc + c^2 = 2rc + c^2. Since we already know that p does not equal q, we can say that 2pc + c^2 does not have to equal 2rc + c^2 - in fact they're only equal if c=0. |
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| OG - PS - #247 |
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