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# Help me in this arithmetic sequence problem

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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?

Oct 8, 2020

#1
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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

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

N = 18 or N - 14 = -4

N = 10 hours      or      N = 18[discard]

Oct 8, 2020
#2
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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 !)

Oct 8, 2020