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The infinite (serial) sequence \(a_1, a_2,\cdots, a_n\)

Expert replies
by AAPL » Fri Aug 16, 2019 8:17 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

Veritas Prep

The infinite (serial) sequence \(a_1, a_2,\cdots, a_n, \cdots\), is such that \(a_1=x, a_2=y, a_3=z, a_4=3\) and \(a_n=a_{n-4}\) for \(n>4\). What is the sum of the first 98 terms of the sequence?

1) \(x=5\)
2) \(y+z=2\)

OA E
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Source: — Data Sufficiency |

by Ian Stewart » Sat Aug 17, 2019 5:33 am
I can't begin to guess what they mean by a "(serial) sequence". A "sequence" is an ordered list of numbers; a "series" is a sum of a sequence. A "(serial) sequence" is not a thing in math.

Here, we know the first four terms are:

x, y, z, 3

and if, after the 4th term, a_n = a_(n-4), then the 5th term equals the 1st, the 6th equals the 2nd, and so on. So the first four terms repeat forever:

x, y, z, 3, x, y, z, 3, x, y, z, 3, ....

We want to sum the first 98 terms. Since 96 = (4)(24), when we write out the first 98 terms, we'll have 24 complete 'blocks' of the first four terms, x, y, z, and 3, along with two additional terms, the 97th and 98th terms, which will be x and y. So the sum will be 24(x + y + z + 3) + x + y. Using both Statements, we know what x+y+z+3 is equal to, and we know what x is equal to, but we don't know what y is equal to (we only know y + z = 2, which does not let us find y), so we cannot evaluate the sum and the answer is E.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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