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The first three terms of an infinite sequence are 2, 7, and 22. After
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21 Jul 2015, 02:08
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The first three terms of an infinite sequence are 2, 7, and 22. After
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Updated on: 21 Jul 2015, 03:07
Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. Following the rule of Infinite sequence we obtain the terms of the sequence as mentioned below 02, 07, 22, 67, 2 02, 6 07, 18 22, 54 67... Observe the last two digits of the sequence which have the cyclicity of 4 and the last two digits repeat in the order {02, 07, 22, 67} Also 35th Term = 4*8 + 3i.e. 35th Terms will have same Last two digits as 3rd term of the sequence = 22 i.e. Sum of lat two digits of 35th Term = 2+2 = 4 Answer: Option B
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Originally posted by GMATinsight on 21 Jul 2015, 02:57.
Last edited by GMATinsight on 21 Jul 2015, 03:07, edited 1 time in total.



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Re: The first three terms of an infinite sequence are 2, 7, and 22. After
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21 Jul 2015, 03:06
Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. Writing the first few numbers in the sequence we get, 2, 7, 22, 67, 202, 607, 1822, 5467... The repeated tens and unit digit seuqneces are 02,07,22,67. Thus the 35th term (8 sequences +3rd term ) will have 22 as its tens and unit's digit. Thus the sum = 2+2 = 4, B is the correct answer.



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Re: The first three terms of an infinite sequence are 2, 7, and 22. After
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22 Jul 2015, 22:17
GMATinsight wrote: Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. Following the rule of Infinite sequence we obtain the terms of the sequence as mentioned below 02, 07, 22, 67, 2 02, 6 07, 18 22, 54 67... Observe the last two digits of the sequence which have the cyclicity of 4 and the last two digits repeat in the order {02, 07, 22, 67} Also 35th Term = 4*8 + 3i.e. 35th Terms will have same Last two digits as 3rd term of the sequence = 22 i.e. Sum of lat two digits of 35th Term = 2+2 = 4 Answer: Option B can this que be solved using formula of sequences in some way..?



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The first three terms of an infinite sequence are 2, 7, and 22. After
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22 Jul 2015, 22:24
riyazgilani wrote: GMATinsight wrote: Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. Following the rule of Infinite sequence we obtain the terms of the sequence as mentioned below 02, 07, 22, 67, 2 02, 6 07, 18 22, 54 67... Observe the last two digits of the sequence which have the cyclicity of 4 and the last two digits repeat in the order {02, 07, 22, 67} Also 35th Term = 4*8 + 3i.e. 35th Terms will have same Last two digits as 3rd term of the sequence = 22 i.e. Sum of lat two digits of 35th Term = 2+2 = 4 Answer: Option B can this que be solved using formula of sequences in some way..? The question is "What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?"The concept of Unit and Ten's digit is bases on CYCLICITY Principle so the apt principle to use here is Cyclicity Principle. However, You may consider that the given sequence is a blend of AP and GP and is called AGP but calculating 35th Term would be too scary as the Common Ratio is 3 which is too big and would finally make the 35th Term as multiple of \(3^{34}\). Hence we, in any case, will never try to calculate 35th Term for this question and will strictly focus our calculations on the last two digits of 35th Term which is already calculated by the method mentioned above. I hope it helps!
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Re: The first three terms of an infinite sequence are 2, 7, and 22. After
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26 Jul 2015, 11:06
Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. 800score Official Solution:(To understand units and tens, in 75 7 is tens and 5 is units.) We cannot reasonably be expected to write the sequence to the thirtyfifth term. We should therefore expect a repeating pattern within the tens and units digits of the sequence. Let’s write out the first 8 terms and see what the pattern is: 2, 7, 22, 67, 202, 607, 1822, 5467. Examining the tens and units digits, we see that the following fourterm pattern repeats in those digits: 02, 07, 22, 67, etc. To find what the tens and units digits of the thirtyfifth term will be, we must first divide the term number (35) by the number of terms in the repeating sequence (4): 35/4 = 8 remainder 3. This means the fourterm sequence fully repeats 8 times, and the remainder tells us how many terms into the repeating sequence the term in question will be. Three terms into the repeating sequence, the tens and units digits are both 2. The sum of these digits is the answer to the question: 2 + 2 = 4. The correct answer is choice (B).
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The first three terms of a sequence are 2, 7, and 22. After the first
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15 Aug 2015, 06:59
The first three terms of a sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. If the sequence continues to be expanded, what will be the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13



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Re: The first three terms of an infinite sequence are 2, 7, and 22. After
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15 Aug 2015, 08:45
Mascarfi wrote: The first three terms of a sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. If the sequence continues to be expanded, what will be the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13 Please search for question before posting.



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The first three terms of an infinite sequence are 2, 7, and 22. After
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15 Aug 2015, 19:20
Bunuel wrote: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirtyfifth term in the sequence?
A. 2 B. 4 C. 7 D. 9 E. 13
Kudos for a correct solution. I did in a similar but slightly different way: 1:0002 2:0007 3:0022 4:0067 5:0202 6:0607 7:1822
At this point I noticed that on the even terms the units digit was 7 and the odd terms the units digit was 2. I now knew on the 35th term the units digit was 2 I then also noticed that the tens digit followed the pattern 0, 0, 2, 6, so knew that on the 36th term the tens digit was 6 therefore the 35th terms tens digit was also 2. 2 + 2 = 4 Hope this slightly different methodology helps someone as much as it does for me to write it out. haha
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Re: The first three terms of an infinite sequence are 2, 7, and 22. After
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14 Jul 2018, 22:55
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