Step-by-step explanation:
we assume day and night are equally long.
and the mouse will start at night of day 1.
then, on day 2, it will first retreat 2 meters (during daylight) and then advance 14 meters (at night).
so, on every full day, the mouse will effectively advance 12 meters (-2 + 14 = 12).
with a starting value of 14 (the first night).
so, we are getting an arithmetic sequence
an = an-1 + c
c = 12 in our case.
a1 = 14
a2 = a1 + 12 = 14+12 = 26
a3 = a2 + 12 = a1 + 12 + 12 = 14 + 2×12 = 38
an = a1 + (n-1)×12 = 14 + 12(n-1)
at what n will it reach 86 ?
86 = 14 + 12(n-1) = 14 + 12n - 12 = 2 + 12n
84 = 12n
n = 7
so, on the 7th day the mouse will reach the surface (if we truly count the first starting night day 1 - this is up to your teacher, but I would).
