Question

If the sum of the even integers between $1$ and k, inclusive, is equal to $2k$, what is the value of k?

• A. $6$
• B. $3$
• C. $2$
• D. $-6$

Answer 1

Answer: (A)

$6$

The sum of the first 'n' even integers is .

2 + 4 + 6 +....+ 2n = n(n + 1)

in this case, k = 2n, $n=\frac{k}{2}$, so the sum of the even integers between 2 and k inclusive is $\frac{k}{2}(\frac{k}{2}+1)=2k$

Solving for k:

$\frac{k^2}{4}+\frac{k}{2}=2k$

$k^2+2k=8k$

$k^2-6k=0$

k(k – 6) = 0

K = 0 (which can't be because k > 2...)

k = 6