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04 Dec 2020, 05:30
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E
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Difficulty:
75%
(hard)
Question Stats:
59% (03:08) correct
41%
(02:45)
wrong
based on 99
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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
Are You Up For the Challenge: 700 Level Questions
A. 4,852
B. 4,951
C. 5,050
D. 490100
E. 500,050
Are You Up For the Challenge: 700 Level Questions
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 ?
S1 = (1)
S2 = (2,3)
S3 = (4,5,6)
S4 = (7,8,9,10)
What is the sum of terms in the set S100 ?
04 Dec 2020, 06:20
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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
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
04 Dec 2020, 10:40
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Bunuel
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.
