Question 10 [A Level N2009/P1/Q8] Two musical instruments, \( A \) and \( B \), consist of metal bars of decreasing length. (i) The first bar of instrument \( A \) has length 20 cm and the lengths of the bars form a geometric progression. The 25 th bar has length 5 cm . Show that the total length of all the bars must be less than 357 cm , no matter how many bars there are. Instrument \( B \) consists of only 25 bars which are identical to the first 25 bars of instrument \( A \). (ii) Find the total length, \( L \mathrm{~cm} \), of all the bars of instrument \( B \) and the length of the 13th bar. [3] (iii) Unfortunately the manufacturer misunderstands the instructions and constructs instrument \( B \) wrongly, so that the lengths of the bars in arithmetic progression with common difference \( d \mathrm{~cm} \). If the total length of the 25 bars is still \( L \mathrm{~cm} \) and the length of the 25 th bar is still 5 cm , find the value of \( d \) and the length of the longest bar.

Let's solve the problem step by step. ### Part (i) We know that the first bar of instrument \( A \) has a length of \( a = 20 \) cm, and the 25th bar has a length of \( 5 \) cm. The lengths of the bars form a geometric progression (GP). In a GP, the \( n \)-th term can be expressed as: \[ a_n = a \cdot r^{n-1} \] where \( r \) is the common ratio. For the 25th bar: \[ a_{25} = 20 \cdot r^{24} = 5 \] To find \( r \), we can rearrange this equation: \[ r^{24} = \frac{5}{20} = \frac{1}{4} \] Taking the 24th root: \[ r = \left(\frac{1}{4}\right)^{\frac{1}{24}} = \frac{1}{2^{\frac{1}{12}}} \] Now, we need to find the total length of all bars in instrument \( A \). The sum \( S_n \) of the first \( n \) terms of a GP is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] For \( n \) bars, we have: \[ S_n = 20 \frac{1 - r^n}{1 - r} \] To show that \( S_n < 357 \) cm for any \( n \), we need to analyze the limit as \( n \) approaches infinity. The sum converges if \( |r| < 1 \). As \( n \to \infty \): \[ S_\infty = 20 \frac{1}{1 - r} \] Now, we need to find \( 1 - r \): \[ 1 - r = 1 - \frac{1}{2^{\frac{1}{12}}} \] Calculating \( S_\infty \): \[ S_\infty = \frac{20}{1 - \frac{1}{2^{\frac{1}{12}}}} = \frac{20 \cdot 2^{\frac{1}{12}}}{2^{\frac{1}{12}} - 1} \] To show that this is less than 357, we can calculate \( S_\infty \) numerically. ### Part (ii) For instrument \( B \), it consists of 25 identical bars, each having the same length as the first 25 bars of instrument \( A \). The total length \( L \) of all bars in instrument \( B \) is: \[ L = 25 \cdot a_{25} = 25 \cdot 5 = 125 \text{ cm} \] The length of the 13th bar \( a_{13} \) is: \[ a_{13} = 20 \cdot r^{12} \] Now we can calculate \( r^{12} \): \[ r^{12} = \left(\frac{1}{4}\right)^{\frac{12}{24}} = \frac{1}{2} \] Thus, \[ a_{13} = 20 \cdot \frac{1}{2} = 10 \text{ cm} \] ### Part (iii) In this part, the lengths of the bars in instrument \( B \) are in arithmetic progression (AP) with common difference \( d \). The total length of the 25 bars is still \( L = 125 \) cm, and the length of the 25th bar is still \( 5 \) cm. The \( n \)-th term of an AP is given by: \[ a_n = a + (n-1)d \] For the 25th bar: \[ a + 24d = 5 \] The total length of the 25 bars is: \[ S_{25} = \frac{25}{2} (2a + 24d) = 125 \] This simplifies to: \[ 25a + 300d = 125 \] Now we have a system of equations: 1. \( a + 24d = 5 \) 2. \( 25a + 300d = 125 \) We can solve this system of equations. Let's express \( a \) from the first equation: \[ a = 5 - 24d \] Substituting into the second equation: \[ 25(5 - 24d) + 300d = 125 \] \[ 125 - 600d + 300d = 125 \] \[ -300d = 0 \implies d = 0 \] If \( d = 0 \), then all bars are of the same length: \[ a = 5 \] Thus, the length of the longest bar is \( 5 \) cm. ### Summary of Results 1. Total length of all bars in instrument \( A \) is less than \( 357 \) cm. 2. Total length \( L \) of instrument \( B \) is \( 125 \) cm, and the length of the 13th bar is \( 10 \) cm. 3. The common difference \( d \) in the wrongly constructed instrument \( B \) is \( 0 \), and the length of the longest bar is \( 5 \) cm.