TRICKY!! sum of even integers!

ser 2,4,6.....(k-1)

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

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

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

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....

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