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.

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### 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.
**Part (i):** - The total length of all bars in instrument $$ A $$ is less than 357 cm, regardless of the number of bars. **Part (ii):** - Total length $$ L $$ of instrument $$ B $$ is 125 cm. - Length of the 13th bar is 10 cm. **Part (iii):** - The common difference $$ d $$ is 0 cm. - The longest bar is 5 cm long.

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