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# The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)

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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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02 Apr 2018, 01:07
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64% (03:00) correct 36% (03:05) wrong based on 88 sessions

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The infinite sequence $$a_1$$, $$a_2$$,… $$a_n$$, … is such that $$a_1 = 4$$, $$a_2 = -2$$, $$a_3 = 6$$, $$a_4 = -1$$, and $$a_n = a_{n-4}$$ for $$n > 4$$. If $$T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}$$, what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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02 Apr 2018, 05:36
1
4
Bunuel wrote:
The infinite sequence $$a_1$$, $$a_2$$,… $$a_n$$, … is such that $$a_1 = 4$$, $$a_2 = -2$$, $$a_3 = 6$$, $$a_4 = -1$$, and $$a_n = a_{n-4}$$ for $$n > 4$$. If $$T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}$$, what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

First figure out what the sequence looks like. Since $$a_n = a_{n-4}$$, $$a_5 = a_1 = 4$$, $$a_6 = a_2 = -2$$, $$a_7 = a_3 = 6$$ and so on...
4, -2, 6, -1, 4, -2, 6, -1, 4, -2, 6, -1, 4, ...

The first 4 terms are repeated. The sum of these terms is 4 - 2 + 6 - 1 = 7

From 10th to 85th terms, we have 85 - 10 + 1 = 76 terms
This means we will have 76/4 = 19 complete sets of 4 terms each so a total sum of 19*7 = 133

Note that it doesn't matter with which term we start. Since we have 76 (a multiple of 4) total terms, we will get 19 complete sets. So if we start with -2, we will end with 4.
-2, 6, -1, 4, -2, 6, -1, ... 4, -2, 6, -1, 4
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Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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02 Apr 2018, 01:34
1
Bunuel wrote:
The infinite sequence $$a_1$$, $$a_2$$,… $$a_n$$, … is such that $$a_1 = 4$$, $$a_2 = -2$$, $$a_3 = 6$$, $$a_4 = -1$$, and $$a_n = a_{n-4}$$ for $$n > 4$$. If $$T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}$$, what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

from 10 to 85 total 19 grps can be formed ,each grp 4 values

a10=a6=a2=-2
simlarly
a11=6,
a12=-1
a13=4
Total sum of one grp=7
Total sum for 19 grps=7*19=133

Option E
Math Expert
Joined: 02 Sep 2009
Posts: 56300
Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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02 Apr 2018, 02:56
Bunuel wrote:
The infinite sequence $$a_1$$, $$a_2$$,… $$a_n$$, … is such that $$a_1 = 4$$, $$a_2 = -2$$, $$a_3 = 6$$, $$a_4 = -1$$, and $$a_n = a_{n-4}$$ for $$n > 4$$. If $$T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}$$, what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

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Posts: 483
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The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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07 Jun 2019, 15:15
Bunuel wrote:
The infinite sequence $$a_1$$, $$a_2$$,… $$a_n$$, … is such that $$a_1 = 4$$, $$a_2 = -2$$, $$a_3 = 6$$, $$a_4 = -1$$, and $$a_n = a_{n-4}$$ for $$n > 4$$. If $$T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}$$, what is the value of T?

(A) 119
(B) 120
(C) 121
(D) 126
(E) 133

Veritas Prep Official Solution

We know the first four terms: a1=4, a2=−2, a3=6, a4=−1

Also it is given that an=an−4an=an−4 i.e. the nth term is equal to the (n-4)th term e.g. 5th term is equal to the 1st term. 6th term is equal to the 2nd term. 7th term is equal to the 3rd term etc.

Hence, the sequence becomes: 4, -2, 6, -1, 4, -2, 6, -1, 4, -2, 6, -1 … (It is always helpful to write down the first few terms of the sequence. It helps you see the pattern.)

The sequence has a cyclicity of 4 i.e. the terms repeat after every 4 terms (go back to the definition of sequence above – it says that the same element can appear multiple times at different positions). Therefore, first to fourth terms will form the first cycle, fifth to eighth terms will form the second cycle, ninth to twelfth terms will form the third cycle and so on…
The sum of each group of 4 terms = 4 – 2 + 6 – 1 = 7

What will be the tenth term, a10? A new cycle starts from a9 so a9=4. Then, a10 must be -2.

a10+a11+a12 is the sum of last three terms of a cycle so this sum must be – 2 + 6 – 1 = 3

a13 to a16 is a complete cycle, a17 to a20 is another complete cycle and so on… The sum of each of the complete cycles is 7.

How many such complete cycles will there be? The first complete cycle will end at a16a16, the second one at a20a20, the third one at a24a24 etc (i.e. at multiples of 4). The last complete cycle will end at a(84). How many complete cycles do we have here then?

16 = 4*4 and 84 = 4*21 so you start from the fourth multiple to the 21st multiple i.e. you have (21 – 4 + 1) = 18 total cycles. If you are confused about the ‘+1’ here, hang on – I will take it up at the end of this post.

The sum of these 18 cycles will be 7*18 = 126 (I know the multiplication table of 18 as should you!)

We still haven’t accounted for a(85), which will be the first term of the next cycle. The first term is 4.

a10+a11+a12+…a84+a85=3+126+4=133=T

or you could just consider this:

You have 18 complete cycles except for the first 3 terms and the last term of the sequence. The last term of the sequence is the first term of a cycle and the first three terms of the sequence are the last three terms of the cycle. So these four terms make one complete cycle. Therefore, instead of 18, you have 19 complete cycles.

T=7∗19=133

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Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)  [#permalink]

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08 Jun 2019, 11:32
1
1
This is a sequence problem where, it’s really not hard to identify the pattern. And once the pattern is established, it’s just a matter of time to get the final answer.

We know $$a_1$$ = 4, $$a_2$$ = -2, $$a_3$$ = 6, $$a_4$$ = -1 and $$a_n$$ = $$a_{n-4}$$ for n= 5, 6, 7,…………..

Plugging in values, we see that,

$$a_5$$ = 4, $$a_6$$ = -2, $$a_7$$ = 6 and $$a_8$$ = -1.

Similarly, $$a_9$$ = 4, $$a_{10}$$ = -2, $$a_{11}$$ = 6 and $$a_{12}$$ = -1.

So, in every cycle, we see that the sum of the terms is 7. This should be a clue to eliminate values which are not multiples of 7. Options B and C can be eliminated.

We are starting with $$a_{10}$$=-2, which is not the start of a cycle. So, should we worry? Not at all, because, we are also ending with $$a_{85}$$ = 4, which is not the end of a cycle. Therefore, we can safely say that we will still be considering the same cycle, albeit with different start and end points.

There are a total of 76 terms from $$a_{10}$$ to $$a_{85}$$ (both inclusive, the number of terms is = 85 -10 +1). This means there are 19 cycles. In each cycle, the sum of the terms is 7. Therefore, the sum of 19 cycles will be 133.

The correct answer option is E.

This is definitely a question which can take about 2 minutes to solve, considering that you have to establish a pattern to solve it, although it falls in the moderately difficult category of questions on sequences. Also, conceptually, there’s nothing challenging in this question; we have mostly solved this question using logic.

Hope this helps!
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Re: The infinite sequence a(1), a(2),… a(n),… is such that a(1) = 4, a(2)   [#permalink] 08 Jun 2019, 11:32
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