Porisma

Worked solution

Compare geometric and arithmetic instrument bars

Instruments A and B have bars of decreasing lengths.

(i) A has first bar 2020 cm and a GP of lengths; its 25th bar is 55 cm. Show the total length is less than 357357 cm, however many bars there are.

B has the first 25 identical bars from A. (ii) Find its total LL and its 13th bar length.

(iii) A mistaken B has 25 AP lengths, the same total LL, and last bar still 55 cm. Find its common difference dd and longest bar.

[2009/I/8]

CJC · Discussion · Question 16

Answer

r=2−1/12r=2^{-1/12}; S∞<357S_\infty<357 cm; L≈272.2573L\approx272.2573 cm; 13th bar 1010 cm; full-precision d≈−0.49085769d\approx-0.49085769 cm and longest bar 16.816.8 cm (3 s.f.).

(i) Bound every finite total by the infinite sum

20r24=520r^{24}=5 and the positive decreasing lengths give r=2−1/12r=2^{-1/12}. Thus S∞=20/(1−2−1/12)≈356.34<357S_\infty=20/(1-2^{-1/12})\approx356.34<357 cm. Every finite total is smaller than this positive infinite sum.

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(ii) Compute the finite length before rounding

L=20(1−r25)/(1−r)≈272.25730618L=20(1-r^{25})/(1-r)\approx272.25730618 cm, giving the printed 272.26272.26 cm or 272272 cm to the nearest centimetre. The 13th bar is 20r12=1020r^{12}=10 cm exactly.

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(iii) Retain full precision and compare the printed rounded calculation

Let aa be the longest AP bar. Then a+24d=5a+24d=5 and 25(a+5)/2=L25(a+5)/2=L, so a=2L/25−5≈16.78058449a=2L/25-5\approx16.78058449 cm and d=(5−a)/24≈−0.49085769d=(5-a)/24\approx-0.49085769 cm. The longest bar is 16.816.8 cm to 3 significant figures. The printed solution substitutes the rounded L=272L=272, giving a=16.76a=16.76, d=−0.49d=-0.49; those intermediate values describe the rounded-total version rather than exactly preserving the original GP total.

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Common mistakes

Prematurely rounding L changes the simultaneous equations in the next part.

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