V5- Sequence

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V5- Sequence

by oquiella » Wed Nov 04, 2015 7:09 am
This is a sequence problem with a and n is right below it. Please repost with the right formatting if possible. I am not sure how to do it. See problem below.


In the sequence a1,a2,a3,..., an,an is determined for all values of n>2 by taking the average of all terms a1 through an−1. If a1=1, what is the value of a7?

(1) a2=19
(2) a3=10


[/img]

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by DavidG@VeritasPrep » Wed Nov 04, 2015 8:28 am
oquiella wrote:This is a sequence problem with a and n is right below it. Please repost with the right formatting if possible. I am not sure how to do it. See problem below.


In the sequence A1,A2, A3..., An, An is determined for all values of n>2 by taking the average of all terms A1 through An-1. If A1=1, what is the value of A7?

(1) A2=19
(2) A3=10


[/img]
Statement 1 tell us that A2=19. If each term (after the second term) is determined by taking the average of all terms A1 through An-1, then we know that A3 is the average of the first two terms, or (1+19)/2 = 10. Similarly, we know that A4 is the average of the first three terms, or (1 + 19 + 10)/3 = 10, and so forth. So our sequence will look like this: 1, 19, 10, 10, 10, 10, 10 .... Clearly, we can find A7, so sufficient.

Statement 2 tell us that A3=10. Well, we know that this term is the average of the two previous terms. So we know that the two previous terms must have summed to 20 to give us an average of 10 for the third term, as 20/2 =10. Thus, we have the same info as statement 1, and we can find A7[/size} This is also Sufficient.

Answer is D
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by vrn2vw » Wed Nov 04, 2015 8:35 am
Alright well I'm not sure how to exactly create the right formatting for an but I do think the OA is D

Statement 1 gives you a1 and a2, from there you can find a3 (a1+a2/2). With a3 you can find a4 ((a1+a2+a3)/3) and you can continue this pattern until you find a7. SUFFICIENT

Statement 2 gives you a1 and a3, which may not seem like enough, but you can find a2 with it. a3 = (a1+a2)/2 --> 10 = (1+a2)/2 --> a2 = 19. With a1, a2, and a3 you can find each number including a7 the same way we did in statement 1. SUFFICIENT

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by oquiella » Wed Nov 04, 2015 3:07 pm
DavidG@VeritasPrep wrote:
oquiella wrote:This is a sequence problem with a and n is right below it. Please repost with the right formatting if possible. I am not sure how to do it. See problem below.


In the sequence A1,A2, A3..., An, An is determined for all values of n>2 by taking the average of all terms A1 through An-1. If A1=1, what is the value of A7?

(1) A2=19
(2) A3=10


[/img]
Statement 1 tell us that A2=19. If each term (after the second term) is determined by taking the average of all terms A1 through An-1, then we know that A3 is the average of the first two terms, or (1+19)/2 = 10. Similarly, we know that A4 is the average of the first three terms, or (1 + 19 + 10)/3 = 10, and so forth. So our sequence will look like this: 1, 19, 10, 10, 10, 10, 10 .... Clearly, we can find A7, so sufficient.

Statement 2 tell us that A3=10. Well, we know that this term is the average of the two previous terms. So we know that the two previous terms must have summed to 20 to give us an average of 10 for the third term, as 20/2 =10. Thus, we have the same info as statement 1, and we can find A7[/size} This is also Sufficient.

Answer is D



Hi David,

What confused me was that the problem stated An is determined for all values of n>2 by taking the average of all terms A1 through An-1. Which made me think the average could not calculated until N>2. Can you explain?

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by DavidG@VeritasPrep » Thu Nov 05, 2015 6:53 am
Hi David,

What confused me was that the problem stated An is determined for all values of n>2 by taking the average of all terms A1 through An-1. Which made me think the average could not calculated until N>2. Can you explain?
Your understanding is correct. And you can see why it makes sense to have this rule begin at n = 3. For n = 1, or A1, clearly it doesn't make any sense to have this number be equal to the average of the previous terms, because there are no previous terms! For n = 2, or A2, the only previous terms is A1. If A2 were equal to the average of the previous term, it would then be the same as A1, which would mean that every term in the sequence would be the same, and there'd be no need for the statements at all! So A3 is the first term in the sequence that is generated by taking the average of the previous terms. A3 = [A1 + A2]/2. And A4 = [A1 + A2 + A3]/3, and so forth
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