Consider the sets of consecutive integers {1}, {2, 3}, {4, 5, 6}, {7,

Question

Consider the sets of consecutive integers {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}, ..., where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let  S_n  be the sum of the elements in the  n_{th}  set. What is the value of  S_{100} ?

A. 4,852
B. 4,951 
C. 5,050
D. 490100
E. 500,050

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Consider the sets
S1 = (1)
S2 = (2,3)
S3 = (4,5,6)
S4 = (7,8,9,10)

What is the sum of terms in the set S100 ?

bhupendersingh27 · Accepted Answer

at first sight it can be figured out that answer should be D or E,

Now as the sum of AP = (1st + last) * n /2

In 100th set, number of element will be 100

so n/2 = 50

now 100 consecutive numbers means 50 odd and 50 even, which means sum of 1st and last terms of the set must be an odd number

from option D, if sum is 490100 then sum of 1st + last terms = (490100/50) = 9802 (even not possible)

from option E if sum is 500050, sum of 1st +last term = (500010/50)= 10001 possible

Answer E

Mona2019 · Answer

The number of elements in each set are increasing by 1 element.
The first element of a set will be the sum of number of elements in previous sets +1
So the first element of S100 is 1+2+3....+99 + 1
= 99*100/2. +1 = 4951

S100 has 100 elements starting from 4951, till 5050

So the sum , using AP sum formula = 100/2 (4951+5050)
= 500 050 
IMO E is the answer

Posted from my mobile device

unraveled · Answer

Mona2019  's solution is better.
The sets are:
S1 = (1)
S2 = (2,3)
S3 = (4,5,6)
S4 = (7,8,9,10)
S5 = (7,8,9,10,11)
....

The best part of the question is that it has patterns, however, they are the worst part as well. We can be sure that A, B and C can never be the answers(we know by reverse solving it).
Here, we can either try to figure out either the first element of the set or the last one.
I couldn't find a find way to find the first element so i tried the later and here's how i did.

Each set's last element if divided by that set number results in an increment of 0.5. For example - 
For S(1), multiplication factor f(1) -  \frac{1}{1} = 1 
for S(2), f(2) =  \frac{3}{2} = 1.5 
for S(3), f(3) =  \frac{6}{3} = 2.0 
... so on

So, we now need to find number of increments when we reach the S(100). The increments are obtained by (n-1) where n is the last set upto which we are trying to figure out the increments.

Since there are 100 sets, number of increments = 100 - 1 = 99
Total increment = 99*0.5 = 49.5
Hence, f(50) = total increment + 1 = 49.5 + 1 = 50.5
the last element of S(100), x = 50.5*100 = 5050

Now, we need to find the first element of S(100), which is 5050 - 100 + 1 = 4951
Finally, we are trying to find  S_100  i.e. the sum of 4951, 4952, ... 5050.
The middle terms of S(100) are 5000 and 5001.
Thus, the S(100) = (5000+5001)/2*100 = 5000.5*100 = 500050

Answer E.

Surely, there must be an easier way.

KarishmaB · Answer

Total number of terms in first 99 sets = 1 + 2 + 3 + 4 + ... + 99 = 99*100/2 =4,950 
(This is the only major calculation required)

So 100th set is {4951, 4952, ... 5050}

ext{Sum of the terms here} = 100*\frac{(4951 + 5050)}{2} 
4951 is about 50 less than 5000 and 5050 is 50 more than 5000 so take their sum as 10,000 and know that it is actually 1 more than 10,000

Sum of the terms is a bit MORE THAN 50*10,000 = 500,000

Answer (E)

Adit_ · Answer

The last number in each set = sum of n integers.
Last number in S100 = 100*101/2 = 5050

Sum of all numbers in S100 = Sum of numbers from 4951 to 5050.

= 100/2 
=50 
=50 
=500050.

Answer:  Option E

dp1234 · Answer

Sum of set with 100 no = 100/2   .... a will be 1st non of set and (a+99) will be 100th no of set. 
Thus sum = 100/2(2a+99) = 100a + 4950. 
Thus last 2 digit will be 50. Only E supports that.

E.

Consider the sets of consecutive integers {1}, {2, 3}, {4, 5, 6}, {7,

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GMAT Problem Solving (PS) Questions

Consider the sets of consecutive integers {1}, {2, 3}, {4, 5, 6}, {7,

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More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards