Set S contains consecutive natural numbers from 1 to 100, in increasin

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: C

Please 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

Solution

Given:

• A set, S, which contains all the natural numbers from 1 to 100, in increasing order.
• \(S_1\) denotes the sum of first n even numbers of the set, S.
• \(S_2\) denotes the sum of last n odd numbers of the set, S.
• \(S_1: S_2\) = 16 : 85

To find:

• The value of n

Approach and Working:
\(S_1\) = sum of first n even numbers = 2 + 4 + 6 + 8 + 10 + …. + 2n

• Now, if we take 2 common from all the terms, we get,

o \(S_1 = 2(1 + 2 + 3 + 4 + 5 + …. + n)\)
o The sum of first ‘n’ natural numbers, \((1+ 2 + 3 + 4 + 5 + …. + n) = \frac{n(n+1)}{2}\)
o Which implies, \(S_1 = 2*n*\frac{(n+1)}{2} = n(n+1)\) ………………… (1)

As we have total 100 consecutive numbers in S, out of those 100 numbers, 50 will be even and rest 50 will be odd.

\(S_2\) = sum of last n odd numbers = sum of all odd numbers from 1 to 100 – sum of the first (50 – n) odd numbers

• The sum of first n odd natural numbers = \(n^2\)

o Implies, the sum of all 50 odd numbers from 1 to \(100 = 50^2\) (Since, the total number of odd numbers from 1 to 100 = 50)
• Similarly, the sum of the first (50 – n) odd numbers = \((50 – n)^2\)
• Thus, \(S_2 = 50^2 - (50 – n)^2\) …………………………… (2)

Substituting, (1) and (2), in \(S_1: S_2 = 16 : 85\), we get,

• \(\frac{n(n + 1)}{[50^2 - (50 – n)^2]} = \frac{16}{85}\)

o If we observe the denominator is in the form of \(a^2 – b^2\)
• Implies, \(\frac{[n *(n + 1)]}{[(100 - n) * (n)]}= 16/85\)

o \(\frac{(n + 1)}{(100 - n)} = \frac{16}{85} = \frac{(15 + 1)}{(100 - 15)}\)

Thus, the value of n = 15

Hence, the correct answer is option C.

Answer: C

The sum of the first n even numbers is n*(n+1) and the ratio of S1:S2 = 16: 85 thus S1 has to be a multiple of 16. Substituting the values of n available we get n=15.

[(2+2n)/2]n/[99+(101-2n)/2]n=16/85➡
202n=3030➡
n=15
C

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