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M31-06

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M31-06  [#permalink]

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New post 07 Jun 2015, 09:01
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A
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C
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  15% (low)

Question Stats:

93% (02:27) correct 7% (01:36) wrong based on 30 sessions

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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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Re M31-06  [#permalink]

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New post 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:

\(t_2=t_1+k\);

\(t_3=t_2+k=t_1+2k\);

\(t_4=t_3+k=t_1+3k\);

...

\(t_n=t_1+(n-1)k\);

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: M31-06  [#permalink]

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New post 08 Jun 2015, 05:38
t2 = t1 + k
t3 = t2 + K = t2 = 2K

tn = t1 + (n-1)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 + (n-1)K, above equation becomes

3t1 + 9K = 3( t1 + 3K) = 3 * 8 = 24
option C
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Re: M31-06  [#permalink]

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New post 29 Oct 2015, 06:56
I think this is a poor-quality 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: M31-06  [#permalink]

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New post 29 Oct 2015, 11:40
danjbon wrote:
I think this is a poor-quality 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: M31-06  [#permalink]

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New post 21 Jul 2016, 06:08
I think this is a high-quality question and I agree with explanation. Nice manipulative question
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Re: M31-06  [#permalink]

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New post 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: M31-06  [#permalink]

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New post 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: M31-06  [#permalink]

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New post 04 Sep 2019, 01:33
1
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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Re: M31-06   [#permalink] 04 Sep 2019, 01:33
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