Sum of three consecutive multiples of $4$ is $84$. The middle is

Sum of three consecutive multiples of $4$ is $84$. The middle is
  1. $28$
  2. $24$
  3. $32$
  4. $20$

Solution

$(4k-4) + 4k + (4k+4) = 12k = 84 \Rightarrow k = 7$. Middle $= 28$.

Asked in: IMO

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