Question

A mouse climbs up a well of 86 meters. It advances 14 meters during the night, but during the day it retreats 2. The mouse will reach the surface on the day:

Answer 1

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).