All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Since we have a geometric progression, $p = 128(\frac{2}{3})^{n-1}$

$\mathrm{ln}p = \mathrm{ln}[128(\frac{2}{3})^{n-1}]$

$\mathrm{ln}p = \mathrm{ln}128 + (n-1)\mathrm{ln}\frac{2}{3}$

$\mathrm{ln}p = \mathrm{ln}(2^{7}) + (n-1)(\mathrm{ln}2 - \mathrm{ln}3)$

$\mathrm{ln}p = 7\mathrm{ln}2 + n\mathrm{ln}2 - n\mathrm{ln}3 - \mathrm{ln}2 + \mathrm{ln}3$

$\mathrm{ln}p = (6+n)\mathrm{ln}2 + (1- n)\mathrm{ln}3$

$\therefore, A=1, B=6, C=-1, D=1$

(ii)
$S_{\infty} = \frac{128}{1-2/3}=384$.
Thus, the length cannot be greater than 384cm.

(iii)
$S_{n}~>~380$

$\frac{128[1-(2/3)^{n}]}{1-2/3}~>~380$

$384[1-(2/3)^{n}]~>~380$

$(\frac{2}{3})^{n}~\textless~\frac{1}{96}$

$n \mathrm{ln}\frac{2}{3}~\textless~\mathrm{ln}\frac{1}{96}$

$n~>~\frac{\mathrm{ln}\frac{1}{96}}{\mathrm{ln}\frac{2}{3}}=11.26$

Therefore, 12 pieces must be cut off.

### KS Comments:

Students should not struggle to find out that this is a geometric progression, what follows is normal manipulation. Many mistakes were made in the last three lines of (iii). Some students were confused about how to handle the inequality after applying the natural logarithm.

### 2 Comments

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