If n is a positive odd number and the sum of the even numbers from 1

Question

If n is a positive odd number and the sum of the even numbers from 1 to n is 79*80, n=?

(A) 79

(B) 81

(C) 157

(D) 159

(E) 161

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chetan2u · Answer

Sum of first n even numbers is n(n+1). 
If there are n numbers, where n is odd, the number of even integers =  \frac{(n-1)}{2}

So sum of first (n-1)/2 even numbers =  \frac{n-1}{2}*(\frac{n-1}{2}+1)=79*80 
 (n-1)(n+1)=4*79*80=2*79*2*80=158*160 , so n=159

OR

2+4+……+(n-1) = 2(1+2+3+….+\frac{n-1}{2})

Sum of first (n-1)/2 positive integers is  \frac{\frac{n-1}{2}(\frac{n-1}{2}+1)}{2}=\frac{(n-1)(n+1)}{8} 
So,  2(1+2+3+….+\frac{n-1}{2})=2* \frac{(n-1)(n+1)}{8}= \frac{(n-1)(n+1)}{4}=79*80 
D

sanjitscorps18 · Answer

Treating this as an equally spaced sequence, we are given the sum of 2 + 4 + ....... + n-1 = 79 x 81

Now average of these numbers = (n-1+2)/2 = (n+1)/2

Count of these numbers = (n-1)/2

Sum = Average x Count in sequence

=> 79 x 80 = (n + 1)(n - 1)/4
=> 79 x 80 x 4 = (n + 1)(n - 1)
=> 158 x 160 = (n + 1)(n - 1)
=> n = 159

Option D

SatvikVedala · Answer

Going from options

sum of even integers from 1 to 79 = 2*(1+2+3+....+39) = 39*40

sum of even integers from 1 to 81 = 2*(1+2+3+....+40) = 40*41

sum of even integers from 1 to 157 = 2*(1+2+3+....+78) = 78*79

sum of even integers from 1 to 159 = 2*(1+2+3+....+79) = 79*80 ---> answer

sum of even integers from 1 to 161 = 2*(1+2+3+....+80) = 80*81

-it's Satvik-
I was here, I was solving & I was learning

Akshaynandurkar · Answer

It is all about patterns
check the following patterns 
for n = 3 , even numbers = 2 , sum = 2 = 1*2
for n = 5 , even numbers = 2,4 , sum = 6 = 2*3
for n = 7 , even numbers = 2,4,6 , sum = 12 = 3*4
for n = 9 , even numbers = 2,4,6,8 , sum = 20 = 4*5
so for n=3 , sum is 1*2 that means for 2nd odd number , smaller factor is 1
similarly for 3rd odd number 5, we have smaller factor = 2
therefore for smaller factor = 79 , odd number is 80th

use AP formula ,
80th odd number = 1+(n-1) *2 = 159
answer is D

Purnank · Answer

There is shortcut to this,

Sum of first m even integers = m(m+1)
Now if we compare this with m(m+1) = 79*80
we will get m as 79.
so there are 79 even numbers present in between 1 to n
so from this we can get that n is 159

Answer is D.

bumpbot · Answer

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If n is a positive odd number and the sum of the even numbers from 1

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If n is a positive odd number and the sum of the even numbers from 1

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