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Joined: 02 Sep 2009
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\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ...\) In the sequence above, each term after the first term is equal to the preceding term plus the constant \(k\). If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)? A. 8 B. 12 C. 24 D. 32 E. 72
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Math Expert
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07 Jun 2015, 09:01
Official Solution:
\(t_1, \ t_2, \ t_3, \ ..., \ t_n, \ ...\) In the sequence above, each term after the first term is equal to the preceding term plus the constant \(k\). If \(t_1+ t_3+t_5+t_7=32\), what is the value of \(t_2+t_4+t_6\)?
A. 8 B. 12 C. 24 D. 32 E. 72
According to the stem: \(t_2=t_1+k\); \(t_3=t_2+k=t_1+2k\); \(t_4=t_3+k=t_1+3k\); ... \(t_n=t_1+(n1)k\); Since \(t_1+t_3+t_5+t_7=32\), then \(t_1+(t_1+2k)+(t_1+4k)+(t_1+6k)=32\). So, \(t_1+3k=8\) We need to find the of \(t_2+t_4+t_6=(t_1+k)+(t_1+3k)+(t_1+5k)=3(t_1+3k)\). From above we know that \(t_1+3k=8\), thus \(3(t_1+3k)=3*8=24\).
Answer: C



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Re: M3106
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08 Jun 2015, 05:38
t2 = t1 + k t3 = t2 + K = t2 = 2K
tn = t1 + (n1)K
equation1  > t1+t3+t5+t7 = 32
Replacing individual component with corresponding t1 , equation 1 becomes 4t1 + 12K = 32 t1 + 3K = 8
Now t2 + t4 + t6 = ? replacing individual component with corresponding t1 + (n1)K, above equation becomes
3t1 + 9K = 3( t1 + 3K) = 3 * 8 = 24 option C



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Re: M3106
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29 Oct 2015, 06:56
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Where is it stated that k is a positive integer? Why can't k be 1/2?



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Re: M3106
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29 Oct 2015, 11:40
danjbon wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Where is it stated that k is a positive integer? Why can't k be 1/2? How does it matter whether k is an integer or not? Where in the solution it's mentioned that k is an integer?



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Re: M3106
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21 Jul 2016, 06:08
I think this is a highquality question and I agree with explanation. Nice manipulative question



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Re: M3106
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19 Jun 2018, 15:50
Possible alternative method?: plugging in values such that the rules are fulfilled in the original problem.
Set K=2, and t1=2.
Sequence (starting with t1) = 2,4,6,8,10,12,14,16,(...) each term is the previous +2, so given constraint in the problem is fulfilled.
t1+t3+t5+t7=2+6+10+14 = 32 (condition is fulfilled).
Then, t2+t4+t6 = 4+8+12 = 24, therefore (C) is the answer.



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Re: M3106
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03 Sep 2019, 20:35
This question was a reminder of Don't make simple things complex. If we could just get the equation and the question asked to 2 variables, we can easily solve the 2 linear equations.



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Re: M3106
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04 Sep 2019, 01:33
Clearly, the sequence defined here is an Arithmetic Progression since consecutive terms differ by a constant value. In an Arithmetic progression, Sum of first term and last term = Sum of second term and second last term = Sum of third term and third last term and so on. Therefore, if we have 7 terms i.e. \(t_1\), \(t_2\), \(t_3\), \(t_4\), \(t_5\), \(t_6\), \(t_7\), then, \(t_1\) + \(t_7\) = \(t_2\) + \(t_6\) = \(t_3\) + \(t_5\) = 2*\(t_4\). If the first term is \(t_1\), \(t_2\) = \(t_1\) + k , \(t_3\) = \(t_1\) + 2k, \(t_4\) = \(t_1\) + 3k and so on. Substituting these values in the equation given, we have, 4*\(t_1\) + 12k = 32, which on simplifying gives us, \(t_1\) + 3k = 8. But, \(t_1\) + 3k = \(t_4\). This means that the fourth term in the sequence, \(t_4\) = 8. As per our discussion above, \(t_2\) + \(t_6\) = 2*\(t_4\). Therefore, \(t_2\) + \(t_4\) + \(t_6\) = 3 * \(t_4\) = 3 * 8 = 24. The correct answer option is C. Hope this helps!
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