High School

A ribbon is cut into \( n \) parts. The length of each part increases such that they form a geometric progression.

Given:
- The length of the fourth part of the ribbon is 9 times the length of the second part of the ribbon.

(a) Calculate the common ratio.

(b) If the total length of the ribbon is \( 1,476.2 \, \text{m} \) and the length of the first part is \( 5 \, \text{cm} \), calculate:

(i) the value of \( n \),

(ii) the length, in cm, of the last part of the ribbon.

Answer :

Answer:

Step-by-step explanation:

[tex]Let\ say\ r \ the \common\ ratio:\\a)a_4=r^2*a_2=9*a_2\Longrightarrow\ r=3\ (not\ -3\ since\ parts\ are\ positive)\\\\b)\\\displaystyle \sum_{i=1}^{n} a_i =a_1+a_2+a_3+...+a_n\\\\=a_1+a_1*r+a_1*r^2+a_1*r^3+...+a_1*r^{n-1}\\\\=a_1*(1+r+r^2+...+r^{n-1})\\\\=a_1*\dfrac{r^n-1}{r-1} \\\\147620=5*\frac{(3^n-1)}{3-1} \\\\3^n-1=\frac{2*147620}{5} \\\\3^n=59049 \\\\n*ln(3)=ln(59049)\\\\n=\dfrac{ln(59049)}{ln(3)} \\\\n=10\\\\a_n=5*3^{10-1}=5*19683=98415(cm)[/tex]

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