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In the sequence above, each term after the first term is equal to the
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13 May 2015, 03:47
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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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Re: In the sequence above, each term after the first term is equal to the
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13 May 2015, 06:03
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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Re: In the sequence above, each term after the first term is equal to the
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13 May 2015, 19:11
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
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13 May 2015, 20:59
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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In the sequence above, each term after the first term is equal to the
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18 May 2015, 05:29
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: \(t_2=t_1+k\); \(t_3=t_2+k=t_1+2k\); \(t_4=t_3+k=t_1+3k\); ... \(t_n=t_1+(n1)k\); 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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Re: In the sequence above, each term after the first term is equal to the
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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.



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In the sequence above, each term after the first term is equal to the
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Updated on: 07 Jul 2018, 19:58
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, 18:58.
Last edited by gracie on 07 Jul 2018, 19:58, edited 1 time in total.



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Re: In the sequence above, each term after the first term is equal to the
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03 Jul 2016, 19:27
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 \(t_2+t_4+t_6 = 3 * t_4 = 24\) Answer (C)
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Re: In the sequence above, each term after the first term is equal to the
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06 Jul 2018, 21:54
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: \(t_2=t_1+k\); \(t_3=t_2+k=t_1+2k\); \(t_4=t_3+k=t_1+3k\); ... \(t_n=t_1+(n1)k\); 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. What about the box symbol that is in front of t3, t5 and t4? How are we saying that t3 = box t3?



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Joined: 02 Sep 2009
Posts: 52431

Re: In the sequence above, each term after the first term is equal to the
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06 Jul 2018, 23:29
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: \(t_2=t_1+k\); \(t_3=t_2+k=t_1+2k\); \(t_4=t_3+k=t_1+3k\); ... \(t_n=t_1+(n1)k\); 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. 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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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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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