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Some good questions..Require solution [#permalink]
08 Nov 2009, 12:18
Question Stats:
50% (03:20) correct
50% (01:18) wrong based on 0 sessions
1. A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
A. 5 B. 6 C. 7 D. 8 E. 9
2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80
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Re: Some good questions..Require solution [#permalink]
08 Nov 2009, 12:51
alok2171 wrote: 1. A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
A. 5 B. 6 C. 7 D. 8 E. 9
8 ways Assuming 1 office # of ways with 3 employees = 1 # of ways with 2 employees = 3 total = 4 two office so answer = 8
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Re: Some good questions..Require solution [#permalink]
08 Nov 2009, 13:02
alok2171 wrote: 2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80
a1 = 2 a2 = -3 a3 = 5 a4 = -1 a5=a5-4= a1 a6=a2 and so on and so forth In other words, the 4 numbers given are the only ones in the sequence We need 97 terms total. Therefore, each term will be seen at least 24 times and 4 times 24 = 96 and the first number will be seen 5 times because we need 97 total: 2 x 25 = 50 -3 x 24 = -72 5 x 24 = 120 -1 x 24 = -24 multiply and add and you get 74
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Re: Some good questions..Require solution [#permalink]
08 Nov 2009, 13:04
alok2171 wrote: 1. A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?
A. 5 B. 6 C. 7 D. 8 E. 9 Each of 3 employees has 2 offices to be assigned so the total # of ways is 2^3=8 Answer: D alok2171 wrote: 2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 I guess that a n=a n-4? If so than: a1=a5, a2=6, a3=a7 etc. So we'll have the sequence 2,-3,5,-1, 2,-3,5,-1, 2,-3,5,-1, ... 97 terms. The sum of group of four terms 2,-3,5,-1 is 3. We'll have 97/4=24 full groups of four, plus 97th term 2. The sum would be 24*3+2=74 Answer: B.
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The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80
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chaitu1315 wrote: The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Is An = A (n-4) ? If yes then the answer should be B A5= A (5-4) = A 1 = 2 similarly A 6 = -3 and this sequence will repeat after every 4 terms A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 3 now Sum of first 96 terms = 3 *24 Sum of 97 terms = 72 + 2 = 74
Last edited by cipher on 02 May 2010, 11:54, edited 1 time in total.
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nitishmahajan wrote: chaitu1315 wrote: The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Is An = A (n-4) ? If yes then the answer should be B A5= A (5-4) = A 1 = 2 similarly A 6 = -3 and this sequence will repeat after every 4 terms A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 2 now Sum of first 96 terms = 2 *24 Sum of 97 terms = 72 + 2 = 74 I understood until the point that as after 4 we have repeat of the sequence, you can say 96/4= 24...as sum of 4 terms is 2, so 2*24= 48...from where did 72 come from...can you explain it further....
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chaitu1315 wrote: nitishmahajan wrote: chaitu1315 wrote: The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Is An = A (n-4) ? If yes then the answer should be B A5= A (5-4) = A 1 = 2 similarly A 6 = -3 and this sequence will repeat after every 4 terms A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 2 now Sum of first 96 terms = 2 *24 Sum of 97 terms = 72 + 2 = 74 I understood until the point that as after 4 we have repeat of the sequence, you can say 96/4= 24...as sum of 4 terms is 2, so 2*24= 48...from where did 72 come from...can you explain it further.... my bad again the was 3 and not 3 edited the post ..!
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