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.
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