GMAT Club invites you to test your GRE knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GRE Strategy guide. Tell your friends to get out their scrap paper and start solving! Click here to view contest & prize details This week's question:Sequence S is such that \(S_n = S_{n-1} + \frac{3}{2}\), and \(S_1 = 2\)

Sequence A is such that \(A_n = A_{n-1} - 1.5\), and \(A_1 = 18.5\)

A | B |

The sum of the terms in S from \(S_1 to S_{13}\), inclusive | The sum of the terms in A from \(A_1 to A_{13}\), inclusive |

A) Quantity A is greater.

B) Quantity B is greater.

C) The two quantities are equal.

D) The relationship cannot be determined from the information given.

Please post your answer, along with the explanation, below. Get cracking! **Edit: **Thank you for participating. The correct answer, and the winner will be announced shortly.

SolutionMany students find sequence notation intimidating, but it doesn’t have to be. Let’s rephrase each sequence in normal language.

Sn = Sn – 1 + is just saying that every term in S is equal to the term before it, plus 3/2.

An = An – 1 – 1.5 is just saying that every term in A is equal to the term before it, minus 1.5.

Of course, 3/2 and 1.5 are the same, which is a good clue that there’s probably some simple way to solve this problem without actually summing up a sequence.

Since S1 = 2 and every term in S is just 3/2 greater than the term before it, Sequence S begins like this: 2, 3.5, 5, 6.5, 8, 9.5, 11....

Since A1 =18.5 and every term in A is just 1.5 less than the term before it, Sequence A begins like this: 18.5, 17, 15.5, 14, 12.5, 11, 9.5....

At this point, it looks as though the two sequences in Column A and Column B are going to have a lot of terms in common! Remember, any common elements appearing in both A and B can just be canceled out.

You could just write out all 13 terms for each column, or you could “skip up” to S13 by noting that S13 is just going to be S1 plus 1.5, twelve times (since it takes twelve “jumps” to get from 1 to 13). 2 plus 18 is 20, so S13 = 20.

You can also “skip up” to the final term in A. To get to A13, take A1 and subtract 1.5 twelve times (since it takes twelve “jumps” to get from 1 to 13). 18.5 minus 18 is 0.5.

So,

Column A looks like this: | And Column B looks like this: |

2, 3.5, 5, ... 18.5, 20 | 18.5, 17, 15.5.... 2, 0.5 |

That is, all the terms from 2 to 18.5 are held in common by both sets, so we can safely subtract them out. Here’s what’s left:

The correct answer is A.

The winner of this Week's Challenge is... (drum roll).... linkiswickeddank. Congratulations! Please send me a pm with your shipping address, and choice of Manhattan GRE Guide

Stay tuned for an update on next week's competition

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