5/284. You drop a ball from a height of 0.5 meter. Each curved path has 52% of the height of the previous path.a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1.b. What height will the ball be at the top of the third path?An) - 50 (0.52)* ), 13.52 cmO A(n) - 0.5 (0.52)*-1, 0.14 cmA(n) = 50 (52)-1; 135,200 cmO A(n) = 0.52. (0.5)*-1, 013 cm30590%

Answer:Part a) The rule of the sequence is [tex]A(n)=50(0.52^{n-1})[/tex]Part b) The height of the ball will be [tex]13.52\ cm[/tex]Step-by-step explanation:Part a) Write a rule for the sequence using centimeters. The initial height is given by the term n = 1.we know thatIn a Geometric Sequence each term is found by multiplying the previous term by a constant called the common ratio (r)In this problem we have a geometric sequenceLetn-----> the number of patha1 ----> is the initial heightr -----> the common ratiowe have[tex]a1=0.5\ m=0.5*100=50\ cm[/tex][tex]r=52\%=52/100=0.52[/tex]The rule for the sequence is equal to[tex]A(n)=a1(r^{n-1})[/tex]substitute[tex]A(n)=50(0.52^{n-1})[/tex]Part b) What height will the ball be at the top of the third path?For n=3      substitute in the equation[tex]A(3)=50(0.52^{3-1})[/tex]  [tex]A(3)=50(0.52^{2})[/tex][tex]A(3)=13.52\ cm[/tex]