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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: 47924
In the sequence above, each term after the first term is equal to the  [#permalink]

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13 May 2015, 04:47
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2
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Difficulty:

45% (medium)

Question Stats:

77% (01:47) correct 23% (02:43) wrong based on 179 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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Posts: 295
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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
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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
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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Posts: 47924
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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Re: 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.
So answer is B

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

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Updated on: 07 Jul 2018, 20:58
t1t1, t2t2, t3t3, ..., tntn, ...
In the sequence above, each term after the first term is equal to the preceding term plus the constant k. If t1+t3+t5+t7=32t1+t3+t5+t7=32, what is the value of t2+t4+t6t2+t4+t6 ?

A. 8
B. 12
C. 24
D. 32
E. 72

t1+t3+t5+t7=32
➡4*t1+12k=32
➡t1+3k=8
t1=2
k=2
t2+t4+t6=4+8+12=24
C

Originally posted by gracie on 03 Jul 2016, 19:58.
Last edited by gracie on 07 Jul 2018, 20:58, edited 1 time in total.
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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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Posts: 6
Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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06 Jul 2018, 22:54
1
Bunuel wrote:
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 $$3(t_1+3k)=3*8=24$$.

What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?
Math Expert
Joined: 02 Sep 2009
Posts: 47924
Re: In the sequence above, each term after the first term is equal to the  [#permalink]

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07 Jul 2018, 00:29
sk1007 wrote:
Bunuel wrote:
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 $$3(t_1+3k)=3*8=24$$.

What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?

That was some bug. Edited. Thank you.
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Re: In the sequence above, each term after the first term is equal to the &nbs [#permalink] 07 Jul 2018, 00:29
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