EgmatQuantExpert wrote:
Set S contains consecutive natural numbers from 1 to 100, in increasing order. \(S_1\) and \(S_2\) denote the sum of first n even numbers and last n odd numbers of set S, respectively. What is the value of n, if \(S_1: S_2\) = 16: 85?
Sum of first n natural numbers is \frac{n*(n+1)}{2}.
Sum of first N even natural numbers can be calculated below:
2+4+6+.....+2n
take 2 common
2(1+2+3+....+n)
=n(n+1)
Now, calculating sum of first n even nos is easy.
However, calculating the sum of last n odd numbers is not
99+97+95+93.It can be done but takes a lot of time.
Let us denote the sum of last n odd numbers by So
Given,
\(\frac{Se}{So}=\frac{16}{85}\)
\(\frac{n(n+1)}{so}=\frac{16}{85}\)
\(85*n(n+1)=16*So\)
\(16=2^4\)
this implies, \(85*n(n+1)=2^4*So\)
since 85 is odd, and So must be an integer. therefore RHS must contain atleast \(2^4\)
This implies that n(n+1) should be such that it provides at least \(2^4\)
Checking Answer choices. for n:
A. 5 - \(5*6\). Drop it.
B. 10 - \(10*11\). Drop it
C. 15 - \(15*16\).
Bingo. Should be our answer, but just for precaution, we'll check the rest of the answer choices.D. 20 - \(21*26\). Drop it.
E. 25 - \(26*26\). Drop it.
Hence our answer: CPlease note that we can calculate the sum of last n odd numbers the following way:
99+97+95+93+.......nth odd no
=(100-1)+(100-3)+(100-5)+.......[100-(2n-1)]
=[100*n] - (1+2+3+......nth term(ie 2n-1)....................a
(1+2+3+......nth term(ie 2n-1)
= sum of first 2n natural no - sum of first n even no.
\(= \frac{2n(2n+1)}{2}-n(n+1)\)
\(=n(2n+1)-n(n+1)\)
\(=2n^2 + n - n^2 -n\)
\(=n^2\)
Using a,
we get, sum of last n odd numbers = 100n-n^2