the answer is A

I know the answer is A from the answers but how did you reach to A? Please explain.

mbamaverick wrote: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

The above question is from The Official Guide for GMAT Quantitative Review, 2nd Edition.

S1 = 1 - (1/2)
S2 = (1/2) - (1/3)
...
Sn = 1/n - (1/(n+1)

Sum it up.

you are left with 1 - (1/n+1)

is 1- (1/n+1) > 9/10

or 1/10 > 1/(n+1)
or n +1 >10
or n > 9

In other words, is #terms > 9

(1) says #terms > 10, sufficient
(2) says #terms < 19. Here #terms can be 7, or 17. Insufficient.

Awesome! Thanks

palvarez wrote:

mbamaverick wrote: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

The above question is from The Official Guide for GMAT Quantitative Review, 2nd Edition.

S1 = 1 - (1/2)
S2 = (1/2) - (1/3)
...
Sn = 1/n - (1/(n+1)

Sum it up.

you are left with 1 - (1/n+1)

is 1- (1/n+1) > 9/10

or 1/10 > 1/(n+1)
or n +1 >10
or n > 9

In other words, is #terms > 9

(1) says #terms > 10, sufficient
(2) says #terms < 19. Here #terms can be 7, or 17. Insufficient.

Can someone explain how to sum up the above sequence and come up with 1 - (1/n+1)?

Sum = ( First term + last term /2) * n { formula to find sum when the enitities are in a sequence)

= 1 + 1/ n ( n+1)

= n ( n+1) +1 / 2 * ( n+1) --- eqn 1



St 1 : K > 10

So minimum K = 11

Put K = 11 in equation 1 :

(11 * 12 +1) /2 * ( 11+1)

133/24

133/24 is > 9/10

Sufficient

St 2:

K < 19

let K = 1 ; Sum = n ( n+1) +1 / 2 * ( n+1)
= 3/4 = 0.75

We have sum < 9/10

Let K = 18 ;Sum = 18 * 19 +1 / ( 2 * ( 18+1)
= 343 / 38
sum is > than 9/10
We have consistent values.
Insufficient

Pick A

silly question--N is equal or greater than 1. What if n is say 10? Then the first term would be 1/10 rather than 1. How does one proceed/simplify then?

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

WRITE IT OUT UNTIL YOU SEE A PATTERN.

S(1) = 1 - 1/2
S(2) = 1/2 - 1/3
S(3) = 1/3 - 1/4
etc.

Sum of the first 3 terms:
S(1) + S(2) + S(3) = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) = 1 - 1/4 = 3/4.

When the terms are added, every value except the first and the last CANCELS out.
The result is that, when n=3, the sum = n/(n+1) = 3/4.
Thus, when n=k, we can DEDUCE that the sum = k/(k+1).

Question rephrased: Is k/(k+1) > 9/10?

Statement 1: k>10.
If k=11, then the sum = k/(k+1) = 11/12, which is GREATER than 9/10.
If k=12, then the sum = k/(k+1) = 12/13, which is GREATER than both 9/10 and the preceding sum, 11/12.

As k increases, so does the sum.
Since the LEAST possible sum here = 11/12, the sum will always be GREATER than 9/10.
SUFFICIENT.

Statement 2: k<19.
As shown in statement 1, if k=11, then the sum is GREATER than 9/10.
If k=1, then the sum = k/(k+1) = 1/2, which is LESS than 9/10.

Since the sum in the first case is GREATER than 9/10 and the sum in the second case is LESS than 9/10, INSUFFICIENT.

The correct answer is A.