Question 846708
*[tex \large A_3 = 2 - 4 = -2]
*[tex \large A_4 = -2 - 2 = -4]
*[tex \large A_5 = -4 - (-2) = -2]
*[tex \large A_6 = -2 - (-4) = 2]
*[tex \large A_7 = 2 - (-2) = 4]
*[tex \large A_8 = 4-2 = 2]


The sequence repeats with period 6 with *[tex \large A_1 = A_7 = A_{13} = \ldots]. Since *[tex \large 2014 \equiv 4 \pmod 6], *[tex \large A_{2014} = a_4 = -4].