GRE Weekly Challenge #1

Question

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!

Thank you for participating. The correct answer, and the winner will be announced shortly. Solution Many 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: Column A Column B 20 0.5 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

72 · Answer

13/2

=13/2 *  
=13/2 * 
=13/2 * 22 = 11*13

13/2 *  
= 37 - 18 = 19

--> Ans = D

rojans · Answer

Compare the sum of s1 to s13 with A1 to A13.

First get the total sum of S1 to S13
We know from formula that next term can be get by adding 1.5 in previous term.
S1=2,S2=3.5,S3=5..................S13=20(Using formula of last term T(n)=a+(n-1)D-where a-fist term ,N-number of terms ,D-conman difference.)
then we have sum of S=2+3.5+5.............+18.5+20 (we can get the sum of all terms ,But we don't need to do that .We have to last term only)
Then comes the sum of A,From A1 to A13.
A1=18.5,A2=17,A3=15.5...................A13=.5(Using the same formula.but this is descending series then we change the sign of common difference ,which become now -1.5 )
A13=18.5+(13-1)x-1.5
A13=.5
A1=18.5,A2=17 Then the sum of A1 to A13
18.5+17..........+2+.5
comparing both series sum of S and A.
Sum of S series = 2+..........+18.5+20
Sum of A series=.5+2+.........+18.5
Both series have same values except S has 20 and a has .5.
we can easily figure out that series S has greater sum than series A.
Hence option A is right.

GyanOne · Answer

Sn = 2,  ,  ,  ........., 
An = 18.5,  ,  ,  ,........,

nth term of Sn =   = 1/2 + 3n/2
nth term of An = 18.5 - (1.5)*(n-1)] = 20 - 1.5n

=> Sum to n terms (Sn) = n/2 + 3n(n+1)/4
 Sum to n terms (An) = 20n - 3n(n+1)/4

Putting in the value of n=13 in both the above expressions, we find that 
Sum to 13 terms (Sn) = 143
Sum to 13 terms (An) = 123.5

Therefore (A) is the correct answer - quantity (A) is greater

rchowdhury03 · Answer

S_n = S_{n-1} + 3/2, and S_1 = 2

A_n = A_{n-1} - 1.5, and A_1 = 18.5

S1+S2+..................+S13= 13*2 + (0+1+2+3+....+12)*3/2 = 26 + 78*3/2 = 143
A1+A2+..................+A13=13*18.5 - (0+1+2+3+....+12)*1.5 = 240.5 - 78*3/2 = 123.5

Therefore, The answer is A
Explanation:
S1 = 2
S2 = S1 + 3/2 = 2 + 3/2
S3 = S2 + 3/2 = 2 + 2*3/2
.
.
.
S13 = 2+ 12*3/2

linkiswickeddank · Answer

A

Sn13 > An1

LoHATra · Answer

Correct answer is A.

A = S1 + S2 + S3 + ... + S13 = 2 + (2 + 1*1.5) + (2 + 2*1.5) + ... + (2 + 12*1.5) = 2 + 3.5 + 5 + ... + 20 = 20 + 18.5 + ... + 3.5 + 2
B = A1 + A2 + A3 + ... + A13 = 18.5 + (18.5 - 1*1.5) + (18.5 - 2*1.5) + ... + (18.5 - 12*1.5) = -----------------> 18.5 + 17 + .... + 2 + 0.5

So A > B.

bhavishbalhotra · Answer

S1=2,S2=3.5,S3=5.......S13=21

Sn=13/2(2+21)=23*13/2

A1=18.5,A2=17,A3=15.5......A13=0.5

An=13/2(18.5+0.5)=19*13/2

So Quantity A is greater

DevilDoggNC · Answer

Ans A

1) S1 to S13 -- 2+3.5+5+6.5+8+9.5+11+12.5+14+15.5+17+18.5+20
2) A1 to A13 -- 18.5+17+15.5+14+12.5+11+9.5+8+6.5+5+3.5+2+.5

20 > .5 (all other terms cancel out between Eq 1 and Eq 2) ----> Ans A

sandy923 · Answer

Sn - Sn-1 = 1.5 == d1 (difference in AP) a1= S1
An - An-1 = -1.5 ==d2 a2=S2

Using
Summation formula = n/2 {2a + (n-1)*d}

1 is the answer.

abhi398 · Answer

option A

guanabana · Answer

Column A

Sum = 13/2

= 13/2

= 143

Column B

Sum = 13/2

= 13/2

= 123.5

So A is bigger

Shalabh9 · Answer

Both A and B will have the same values as one'll end up adding the same numbers. The series S has the ascending order of the same series of numbers that are in descending order in the series A.

Thank You.

kingreuben · Answer

Sn = Sn-1 + 3/2 we know S1 = 2 S2 = S1+3/2 S3= S2+3/2 = S1+(S1+3/2)+3/2 summation from 1 to 13 will be S1+S2+.....S13 The sequence will then appear as S1 + (S1+3/2)+(S1+3/2+3/2)+(S1+3/2+3/2+3/2)+... ie each subsequent term add 3/2 to the previous term . 13S1+ 3/2(13-1) 13S1+3/2(1+2+3+4+5.....+12) 13(2)+ x 3/2(78) = 143 similarly the summation 13(18.5) - 3/2(78) = 123.5 clearly A is greater than B. the correct Ans is A :oops:

prashantserai · Answer

A) Quantity A is greater.

Qty A considers sum of S1 through S13
and Qty B considers sum of A1 through A13

Qty A = (2)+(2+1*1.5) +...+(2+12*1.5) = (13* 2) + 1.5(1+2+...12) 
Qty B = (18.5)+(18.5 - 1*1.5) +...+ (18.5-12*1.5) = (13*18.5) - 1.5(1+2+...12)

Now there are two parts to each term of quantity the base number (2 or 18.5) and the addition (+1.5 or -1.5)
In the first aspect Qty B is greater than A by (18.5-2)*13 = 16.5*13 = 160+48+6.5 = 214.5
In the second aspect Qty A is greater than B by (1+2+...12)*(1.5 - (-1.5)) =(6.5*12)*3 = (72+6)*3 = 210+24 = 234
Now since 234>214.5

Qty A is greater than Qty B.

PS: I have shown more steps, but each one of these may be very convenient and instant calculations.

GRE Weekly Challenge #1

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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

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