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A rod one metre in length is divided into $10$ pieces whose lengths are in geometrical progression. The length of the longest piece is eight times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest piece.

I am working through a pure maths book as a hobby. I believe the length of shortest piece = $a$. Length of longest piece $= ar^{9}$, because

$$\frac{ar^{9}}{a} = 8 \implies r^{9} = 8 \implies r^{3} = 2 \implies r = 2^{\frac{1}{3}}$$

$$S_n = \frac{a(r^{10} - 1)}{r - 1} = \frac{a((2^{1/3})^{10} - 1)}{r - 1} = \frac{a(2^{10/3} - 1)}{r -1}$$

$$\implies \frac{9.9079}{4/3} = 1000 \implies a \approx 135$$

but the book says the answer is $29$ mm.

Jessie
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Steblo
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    Where did that $4/3$ come from? $r=2^{1/3}$ is not the same as $2+1/3$. – Jaap Scherphuis Mar 05 '21 at 11:19
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    At some point you decide that the $r$ above is equal to $2^{1/3}$, but that the $r$ below is a story for another time. And also that $a$ must be sent to a re-education camp (except that it then shows up after a while). But other than those last few lines you were fine. –  Mar 05 '21 at 11:22
  • Thanks for these comments. A stupid mistake on my part! – Steblo Mar 05 '21 at 12:17

1 Answers1

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Your mistake is in the last step. It should be:

$S_n =\frac{a \cdot (2^{10/3} - 1)}{2^{1/3} - 1} \approx a \cdot 34.93$

or $a \approx 28.63 \approx 29$

as given in the solution

user0
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