During the first hour of her climb, a mountain climber ascends 675 ft. Each hour after that, she ascends 50ft less than in the preceding hour. After how many hours will she be 4500 ft above her starting point?

Guest Oct 8, 2020

#1**0 **

Use the AP formula to find number of hours.

Sum = N/2 * [2*F + D*(N - 1)]

4500 =N/2 * [2*675 + (-50*(N-1))], solve for N

Solve for N:

4500 = 1/2 N (1350 - 50 (N - 1))

4500 = 1/2 (1350 - 50 (N - 1)) N is equivalent to 1/2 (1350 - 50 (N - 1)) N = 4500:

1/2 N (1350 - 50 (N - 1)) = 4500

Multiply both sides by 2:

N (1350 - 50 (N - 1)) = 9000

Expand out terms of the left hand side:

1400 N - 50 N^2 = 9000

Divide both sides by -50:

N^2 - 28 N = -180

Add 196 to both sides:

N^2 - 28 N + 196 = 16

Write the left hand side as a square:

(N - 14)^2 = 16

Take the square root of both sides:

N - 14 = 4 or N - 14 = -4

Add 14 to both sides:

N = 18 or N - 14 = -4

Add 14 to both sides:

**N = 10 hours ** or N = 18[discard]

Guest Oct 8, 2020

#2**+1 **

hr 1 2 3 4 5 6 7

675 + 625 + 575 + 525 + 475 +425 +375 = 3675 ft Not quite there....add some more hours

hr 7 8 9 10

3675 + 325 + 275 +225 = 4500 ft

(sometimes it is just shorter to 'brute force' it !)

ElectricPavlov Oct 8, 2020