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

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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.
[Reveal] Spoiler: OA

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

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New post 13 May 2015, 07:03
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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

Answer: C

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New post 13 May 2015, 20:11
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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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New post 13 May 2015, 21:59
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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Re: In the sequence above, each term after the first term is equal to the [#permalink]

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New post 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 \(3(t_1+3k)=3*8=24\).

Answer: C.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 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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New post 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

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

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New post 03 Jul 2016, 20:27
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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 \(t_2+t_4+t_6 = 3 * t_4 = 24\)

Answer (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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