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two cities are building bicycle paths. - charles city has built 5 km of…

Question

two cities are building bicycle paths. - charles city has built 5 km of bicycle paths by the end of the first month, and the total length of the paths doubles each month. - tinsel town has built 21 km of bicycle paths by the end of the first month, and the total length of the paths increases by 5 km per month. at the end of which month does the total length of the bicycle paths in charles city first exceed the length in tinsel town?

Explanation:

Step1: Define the growth models

Let \( n \) be the number of months. For Charles City, the length of bicycle paths follows a geometric sequence with initial term \( a = 5 \) and common ratio \( r = 2 \). So the length after \( n \) months is \( C(n)=5\times2^{n - 1}\) (since at the end of the first month, it's 5 km, so for \( n = 1 \), \( C(1)=5\times2^{0}=5\), for \( n = 2 \), \( C(2)=5\times2^{1} = 10\), etc.). For Tinsel Town, the length follows an arithmetic sequence with initial term \( b = 21 \) and common difference \( d=5 \). So the length after \( n \) months is \( T(n)=21 + 5\times(n - 1)=16 + 5n\).

Step2: Calculate lengths for each month

  • Month 1:
  • Charles City: \( C(1)=5\times2^{0}=5\) km.
  • Tinsel Town: \( T(1)=16+5\times1 = 21\) km. \( 5<21 \)
  • Month 2:
  • Charles City: \( C(2)=5\times2^{1}=10\) km.
  • Tinsel Town: \( T(2)=16 + 5\times2=26\) km. \( 10<26 \)
  • Month 3:
  • Charles City: \( C(3)=5\times2^{2}=20\) km.
  • Tinsel Town: \( T(3)=16+5\times3 = 31\) km. \( 20<31 \)
  • Month 4:
  • Charles City: \( C(4)=5\times2^{3}=40\) km.
  • Tinsel Town: \( T(4)=16+5\times4=36\) km. \( 40>36 \)

Answer:

At the end of the 4th month, the total length of the bicycle paths in Charles City first exceeds that in Tinsel Town.