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
ANSWER: E