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TRICKY!! sum of even integers!

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by xcusemeplz2009 » Mon Nov 02, 2009 4:08 am
ser 2,4,6.....(k-1)

sum=((2+k-1)/2)*79
79*80=((1+k)/2)*79
sol
k=159
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by gmatmaths » Mon Nov 02, 2009 4:11 am
Sn=nA1+[n(n-1)d]/2

n=(k-1)/2, A1=2, d=2 ===>

2n+n(n-1)=n(n+1) ===>

[(k-1)/2]*[(k-1+2)/2]=(k-1)*(k+1)/4 ====>

(k-1)*(k+1)=79*2*80*2 ===>

k=159
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by vikram4689 » Mon Nov 02, 2009 5:40 am
sum = total numbers x average

series = 2,4,6,......k-1
total numbers = [(k-1)-2]/2 + 1 = (k-1)/2
average = (k-1+2)/2 = (k+1)/2

sum = (k-1)(k+1)/2*2 = 79*80
k-1 * k+1 = 158 * 160
k = 159

P.S. @xcusemeplz2009,.... how u got total numbers as 79
though the answere is right but i would like to know how u got to know about total numbers = 79
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by xcusemeplz2009 » Mon Nov 02, 2009 6:22 am
vikram4689 wrote:sum = total numbers x average

series = 2,4,6,......k-1
total numbers = [(k-1)-2]/2 + 1 = (k-1)/2
average = (k-1+2)/2 = (k+1)/2

sum = (k-1)(k+1)/2*2 = 79*80
k-1 * k+1 = 158 * 160
k = 159

P.S. @xcusemeplz2009,.... how u got total numbers as 79
though the answere is right but i would like to know how u got to know about total numbers = 79
as the series is of even int and sum is given as 79*80 , the total term will be either 158 or 160(we can conclude this by remembering the formula of finding the sum of an AP ser i.e
total no. of term*(1st term value+last term value)/2

in this case
x*(2+y)/2=79*80

if we take value of y as 160 we get sum as 81*80
for y=158 we get 79*80

since k is odd it has to be 159...

if anyone is aware of the formula for sum of any series in AP then this problem can be solve with out any paperwork....
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by Jeff@TargetTestPrep » Mon Dec 11, 2017 4:16 pm
vkb16 wrote:The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159
The sum of the first k positive even numbers is k(k + 1). For example, the sum of the first 5 positive even numbers is 5(6) = 30. We can verify this by actually adding the first 5 positive even numbers: 2 + 4 + 6 + 8 + 10 = 30. Therefore, to solve this problem, we need to find the number of even numbers between 1 and n, in which n is an odd number. In this range, the smallest even number is 2 and the largest even number is (n - 1) since n is odd. Thus, the number of even numbers between 1 and n is:

[(n - 1) - 2]/2 + 1 = (n - 1)/2 - 1 + 1 = (n - 1)/2

Thus, the the sum of the even numbers between 1 and n is (n - 1)/2 * [(n - 1)/2 + 1].
We are given that this sum is equal to 79*80, so we can set (n - 1)/2 * [(n - 1)/2 + 1] = 79*80 and solve for n.

However, since (n - 1)/2 + 1 is 1 more than (n - 1)/2 and 80 is one more than 79, (n - 1)/2 must be 79 and (n - 1)/2 + 1 must be 80. Now we can determine n:

(n - 1)/2 = 79

n - 1 = 158

n = 159

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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