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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 \(t_2+t_4+t_6\) ?

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

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13 May 2015, 20:59

1

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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

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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\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ... 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 \(3(t_1+3k)=3*8=24\).

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

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03 Jul 2016, 03:58

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.

\(t_1\), \(t_2\), \(t_3\), ..., \(t_n\), ... 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 \(t_4\).

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

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07 Oct 2017, 00:57

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