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Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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Difficulty:   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
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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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 $$3(t_1+3k)=3*8=24$$. Joined: 24 Oct 2012
Posts: 163

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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
Intern  Joined: 14 Oct 2015
Posts: 29
GMAT 1: 640 Q45 V33 ### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

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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?
Senior Manager  Joined: 31 Mar 2016
Posts: 375
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
WE: Operations (Commercial Banking)

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I think this is a high-quality question and I agree with explanation. Nice manipulative question
Intern  B
Joined: 12 Jan 2018
Posts: 17

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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.
Intern  B
Joined: 09 Dec 2018
Posts: 25
Location: India
GMAT 1: 650 Q47 V32 GPA: 3

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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.
CrackVerbal Quant Expert G
Joined: 12 Apr 2019
Posts: 309

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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!
_________________ Re: M31-06   [#permalink] 04 Sep 2019, 01:33
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# M31-06  