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# In the sequence above, each term after the first term is equal to the

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Math Expert
Joined: 02 Sep 2009
Posts: 46035
In the sequence above, each term after the first term is equal to the [#permalink]

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13 May 2015, 04:47
1
2
00:00

Difficulty:

45% (medium)

Question Stats:

77% (01:35) correct 23% (02:39) wrong based on 147 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

Kudos for a correct solution.

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Senior Manager
Joined: 28 Feb 2014
Posts: 295
Location: United States
Concentration: Strategy, General Management
Re: In the sequence above, each term after the first term is equal to the [#permalink]

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13 May 2015, 07:03
2
1
x+x+2k+x+4k+x+6k=32
4x+12k=32
x+3k=8

We are looking for the value of:
x+k+x+3k+x+5k=?
3x+9k=?

3(x+3k=8)
3x+9k=24

Intern
Joined: 22 Mar 2015
Posts: 3
Re: In the sequence above, each term after the first term is equal to the [#permalink]

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13 May 2015, 20:11
1
t1 + t3 +t5 +t7 = t1 + t1 +2k +t1 +4k +t1 +6k= 4t + 12k = 32
t = 2 ==> k =2 ==> t2+t4 +t6 = 4 +8 +12 = 24
=> C
Senior Manager
Joined: 21 Jan 2015
Posts: 283
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28
WE: Sales (Consumer Products)
Re: In the sequence above, each term after the first term is equal to the [#permalink]

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13 May 2015, 21:59
1
Bunuel wrote:
$$t_1$$, $$t_2$$, $$t_3$$, ..., $$t_n$$, ...
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

Ans: C
Solution: according to given condition
t2=t1+k
t3=t2+k = t1+2k
same way t4=t1+3k
t5=t1+4k
t6=t1+5k
t7=t1+6k

now by putting the values in given equation
t1+t3+t5+t7=32
t1+3k=8

required value
t2+t4+t6 = 3(t1+3k)= 3*8= 24

Ans: C
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Math Expert
Joined: 02 Sep 2009
Posts: 46035
Re: In the sequence above, each term after the first term is equal to the [#permalink]

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18 May 2015, 06:29
Bunuel wrote:
$$t_1$$, $$t_2$$, $$t_3$$, ..., $$t_n$$, ...
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

Kudos for a correct solution.

OFFICIAL SOLUTION:

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 $$3(t_1+3k)=3*8=24$$.

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Intern
Joined: 21 Feb 2016
Posts: 9
Location: United States (MA)
In the sequence above, each term after the first term is equal to the [#permalink]

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03 Jul 2016, 04:58
Bunuel wrote:
$$t_1$$, $$t_2$$, $$t_3$$, ..., $$t_n$$, ...
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

Kudos for a correct solution.

According to the question, t2=t1+k, t3=t2+k=t1+k+k So, t3=t1+2k, t4=t1+3k, t5=t1+4k.. That's the pattern.
So, t1+t3+t5+t7
=t1+t1+2k+t1+4k+t1+6k
=4t+12k which is 32 given
At this point, u can take any value of t and k that satisfies the equation 4t+12k=32. (The (t,k) pairs could be (2,2) or (5,1).
Now, we are asked to find t2+t4+t6, which can be written as 3t+k+3k+5k, which in turn is equal to 3t+9k.
Taking either pair of values of (t,k), we get 3t+9k=6+18=24.

Alternatively, you could see that as t1+t3+t5+t7=32, and as 1+3+5+7=16, so the total, to be 32, each term has to be double, i.e. 2+6+10+14=32. Now these terms are not consecutive meaning there is one term between any two consecutive of these terms. So, t2+t4+t6 would be 4+8+12, which is 24.
Hope it helps.
VP
Joined: 07 Dec 2014
Posts: 1018
In the sequence above, each term after the first term is equal to the [#permalink]

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03 Jul 2016, 19:58
let x=value of t2+t4+t6
if t1+t3+t5+t7=32,
then 32/4=x/3
x=24
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8097
Location: Pune, India
Re: In the sequence above, each term after the first term is equal to the [#permalink]

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03 Jul 2016, 20:27
Bunuel wrote:
$$t_1$$, $$t_2$$, $$t_3$$, ..., $$t_n$$, ...
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

Kudos for a correct solution.

Another method is to note that the average of the two sequences would be the same.

The average of $$t_1$$, $$t_3$$, $$t_5$$ and $$t_7$$ will be the middle of $$t_3$$ and $$t_5$$ which is $$t_4$$. (All terms in an AP are equidistant from the term before it and the term after it.)
The average of $$t_2$$, $$t_4$$, $$t_6$$ is obviously $$t_4$$.

$$t_1+t_3+t_5+t_7=4 * t_4 = 32$$

$$t_4 = 8$$

Then $$t_2+t_4+t_6 = 3 * t_4 = 24$$

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Re: In the sequence above, each term after the first term is equal to the [#permalink]

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07 Oct 2017, 01:57
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Re: In the sequence above, each term after the first term is equal to the   [#permalink] 07 Oct 2017, 01:57
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