go 12 #157

[c210]

Hey guys just a little stumped on a question on the OG12 #157

question is :


For any positive integer n, the sum of the first
n positive integers equals n(n +1) . What is the sum

of all the even integers between 99 and 301 ?

the OG12 obviously gives the answer but I can't for the life of me understand why you multiply each equation n(n+1) /2 by 2??

any help would be appreciated!

best,

c

[Bill@VeritasPrep]

We actually just looked at a similar question on the Veritas blog: https://www.veritasprep.com/blog/2012/03 ... gressions/

I solved it slightly differently:

For "What is the sum of all the even integers between 99 and 401?", we could also solve using our knowledge of evenly distributed sets.

We have a set of even numbers from 100 to 400 (since 99 and 401 are odd), so the range is 300. This means that we have (350/2) + 1 = 151 even numbers. (We add one because it is inclusive).

We also know that the median in an evenly distributed set can be found by finding the mean of the minimum and maximum values. 100 + 400 = 500; 500/2 = 250.

The final step is to multiply 151 and 250. 250 * 100 = 25000; another half of that would be 12500 for a sum of 37500. Add 250 for the 1 that I rounded down and we arrive at the answer of 37750.

Bill

[seal4913]

Its used for positve intergers. They r asking for even numbers. However that formula is for all. So to get even number u divide by 2. For example 1 to 10 so evem is 110 divided by 2 which is 5 even numbers

[ronnie1985]

The sum of first n even numbers starting from 2 is given by
S = 2+4+6+...+2n = 2*(1+2+3+...+n) = n*(n+1)
I hope this clears your doubt.

[[email protected]]

We could have also used the formula of arithmetic progression...

(i) Tn = a + (n-1)d where a = 100 ; d = 2 and found out the value of 'n' i.e 150


(ii) the sum of n even numbers...

= n/2 X [2a + (n-1)d] to find out the sum of even numbers...


the answer comes out to be 37750...

hope this helps!