Can someone please explain to me how this problem works? Why do they rewrite the last terms as (1)(x2)?
Rick
Quantative Review 2nd Edition, Problem Solving #133
This topic has expert replies
I actually posted the wrong number it should be #107.
The sequence s1, s2, s3, .... sn.... is such that sn = 1/n  1/n+1 for all integers n >= 1. If K is a positive integer, is the sum of the first K terms of the sequence greater than 9/10?
1. K > 10
2. K < 19
Thanks
The sequence s1, s2, s3, .... sn.... is such that sn = 1/n  1/n+1 for all integers n >= 1. If K is a positive integer, is the sum of the first K terms of the sequence greater than 9/10?
1. K > 10
2. K < 19
Thanks
Sn=1/n  1/n+1
S1=11/2=1/2
S2=1/21/3=1/6
S3=1/31/4=1/12
k=2,S1+S2=2/3
k=3,S1+S2+S3=3/4
So Sk=11/(k+1)
Is Sk>9/10
11/(k+1)>9/10
1/10>1/(k+1)
k+1>10
k>9?
stmt1,
k>10
So surely k>9
Suff
stmt2,
k<19
k can be k<9 or k>9
Not suff
Pick A
S1=11/2=1/2
S2=1/21/3=1/6
S3=1/31/4=1/12
k=2,S1+S2=2/3
k=3,S1+S2+S3=3/4
So Sk=11/(k+1)
Is Sk>9/10
11/(k+1)>9/10
1/10>1/(k+1)
k+1>10
k>9?
stmt1,
k>10
So surely k>9
Suff
stmt2,
k<19
k can be k<9 or k>9
Not suff
Pick A
Anand

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To understand how the sequence can be simplified you can just write out the various terms of the sequence putting in various values for n.
s1 = 1/1  1/2
s2 = 1/2  1/3
s3 = 1/3  1/4
s4 = 1/4  1/5
...
...
s(k1)th term = 1/(k1)  1/k
sk = 1/k  1/(k+1)
Now, if we add them up then we can see that values cancel out; 1/2 and +1/2, 1/3 and +1/3 and so on. The only terms that remain are 1/1 and 1/(k+1).
Therefore,
the problem boils down to deciding whether 1 1/(k+1) is greater than 9/10 or not.
So, the inequality that we should be trying to solve is
1  1/(k+1) > 9/10
Subtract 1 from both sides,
1/(k+1) > 1/10
Multiplying by 1, reverses the sign,
1/(k+1) < 1/10
Now cross multiply; this won't change the sign since k is positive and so is k+1.
10 < k+1
or, 9<k
So, the sum of the sequence will be greater than 9/10 if the value of k is greater than 9.
Statement 1 tells us that k>10. So, k ios definitely greater than 9 which means it guarantees that the sum of the sequence will be greater than 9/10.
Statement 2 tells us that k < 19. Now as long as k > 9 the sum of the sequence will be greater than 9/10 but when the value of k becomes less than 9, the sum will not be greater than 9/10. Hence, with statement 2 we have both possibilities. Thus this statement is not sufficient.
Hope, this helps you understand.
This question got me a little confused. Isn't 'sn' a standard notation supposed to mean the Sum of a sequence upto its nth term? To denote a sequence a simple 's' is used. Now, these are universal notations and GMAAC shouldn't be messing around with such standard notations, unless they explicitly mention something in the line of "there is a sequence that can be expressed as".
Thanks
Anirban
s1 = 1/1  1/2
s2 = 1/2  1/3
s3 = 1/3  1/4
s4 = 1/4  1/5
...
...
s(k1)th term = 1/(k1)  1/k
sk = 1/k  1/(k+1)
Now, if we add them up then we can see that values cancel out; 1/2 and +1/2, 1/3 and +1/3 and so on. The only terms that remain are 1/1 and 1/(k+1).
Therefore,
the problem boils down to deciding whether 1 1/(k+1) is greater than 9/10 or not.
So, the inequality that we should be trying to solve is
1  1/(k+1) > 9/10
Subtract 1 from both sides,
1/(k+1) > 1/10
Multiplying by 1, reverses the sign,
1/(k+1) < 1/10
Now cross multiply; this won't change the sign since k is positive and so is k+1.
10 < k+1
or, 9<k
So, the sum of the sequence will be greater than 9/10 if the value of k is greater than 9.
Statement 1 tells us that k>10. So, k ios definitely greater than 9 which means it guarantees that the sum of the sequence will be greater than 9/10.
Statement 2 tells us that k < 19. Now as long as k > 9 the sum of the sequence will be greater than 9/10 but when the value of k becomes less than 9, the sum will not be greater than 9/10. Hence, with statement 2 we have both possibilities. Thus this statement is not sufficient.
Hope, this helps you understand.
This question got me a little confused. Isn't 'sn' a standard notation supposed to mean the Sum of a sequence upto its nth term? To denote a sequence a simple 's' is used. Now, these are universal notations and GMAAC shouldn't be messing around with such standard notations, unless they explicitly mention something in the line of "there is a sequence that can be expressed as".
Thanks
Anirban