H. V. Hilprecht. Mathematical, Metrological and Chronological Tablets from the Temple Library of Nippur. Univ. of Pennsylvania, Philadelphia, 1906. Pp. 13, 28-34, 62, 69, pl. 15, PL. IX, are about a tablet which has a geometric progression from c-2300. The progression is double: a_n = 125 * 2ⁿ and 60⁴/a_n for n = 0, 1, ..., 7. There is no summation.

Tablet K 90 of the British Museum. A moon tablet deciphered by Hincks containing 5, 10, 20, 40, 80 followed by 80, 96, 112, 128, ..., 240. Described in The Literary Gazette (5 Aug 1854) _ ??NYS. Cited and described in: Nicomachus of Gerasa: Introduction to Arithmetic; Translated by Martin Luther D'Ooge, with notes by Frank Egleston Robbins and Louis Charles Karpinski; Macmillan, London, 1926; p. 12.

Euclid. IX: 35, 36. This gives the general rule for the sum of a geometric progression.

Zhang Qiujian . Zhang Qiujian Suan Jing. Op. cit. in 7.E. 468, ??NYS. Mikami 42 gives: "A horse, halving its speed every day, runs 700 miles in 7 days. What are his daily journeys?" _ i.e. x*(1 + 1/2 + ... + 1/64) = 700. Solved by adding up.

Mahavira. 850.

Chap. IV, v. 28, p. 74: x - x/2 - x/4 - ... - x/256 = 32.

Chap. VI, v. 314, pp. 175-176: Let a_i+1 = r*a_i + c. He sums such terms.

Fibonacci. 1202. Pp. 313-316. Man has 100 and gives away 1/10 of his wealth 12 times. This has been described under 7.E.

Columbia Algorism. c1370. Prob. 63, pp. 84-85. Same as the Fibonacci, but he converts to pounds, shillings and pence!

Lucca 1754. c1390. F. 10v, pp. 36-37. Computes 2⁴⁰ & 2¹⁰⁰ by repeated squaring.

Chuquet. 1484. He gives a number of such problems _ see also 7.E.

Prob. 96, English in FHM 219. Cask of 9½ drains so first barrel takes 1 hour, second barrel takes 2 hours, third barrel takes 4 hours, .... How long to empty? I.e. he wants the sum of 9½ terms of a geometric progression. He gets the correct answer of 2^9.5 - 1 hours.

Prob. 97, English in FHM 219. Man travels 1, 3, 9, ... leagues per day. How far has he travelled in 5½ days? He gets the correct answer of (3^5.5-1)/2 days.

Ghaligai. Practica D'Arithmetica. 1521. Prob. 29, f. 66v. Same as Fibonacci, pp. 313-316. (H&S 59-60.)

Buteo. Logistica. 1559.

Prob. 69, pp. 276-278. 1 + .9 + (.9)² + ... + (.9)^x = 7.5. The phrasing of the problem is unclear, but this is what he considers. He interpolates linearly between 12 and 13, getting 12.164705107, while the exact answer is (log .25/log .9) - 1 = 12.15762696.

Prob. 84, pp. 294-296. Relates 2ⁱ and 5ⁱ for i = 1, ..., 7. He determines 2^3.5 as Ö128 which he estimates as 11_. Similarly he estimates 5^3.5 as 279.

Jacques Chauvet Champenois. Les Institutions de L'Arithmetique. Hierosme de Marnef, Paris, 1578, p. 70. ??NYS. Problem of tailor and robe involving 4888 divided by 2 twenty times. (French quoted in H&S 14-15.)

van Etten. 1624. Prob. 87 (84): Des Progressions & de la prodigieuse multiplication des animaux, des Plantes, des fruicts de l'or & de l'argent quand on va tousjours augmentant par certaine proportion, pp. 111-118 (177-183). Numerous examples including horseshoe problem and chessboard problem, with ratios 1000, 4, 50. Henrion's Notte, p. 38, observes that there are many arithmetical errors which the reader can easily correct. In part X: Multiplication des Hommes, he considers one of the children of Noah, says a generation takes 30 years and that, when augmented to the seventh, one family can easily produce 800,000 souls. The 1674 English ed. has: "... if we take but one of the Children of Noah, and suppose that a new Generation of People begin at every 30 years, and that it be continued to the Seventh Generation, which is 200 years; ... then of one only Family there would be produced 111000 Souls, 305 to begin the World: ... which number springing onely from a simple production of one yearly ...."

Leybourn. Pleasure with Profit. 1694. Chap. VI, pp. 24-28: Of the Increase of Swine, Corn, Sheep, &c. Examples with ratios 4, 40, 2, 1000, 2, mostly taken from van Etten. Then art. VI: Of Men, discusses the repopulation of the world from Noah's children: "... if we take but one of the Children of Noah, and suppose that a New Generation of People begin at every 30 years, and that it be continued to the seventh Generation, which is 210 years; ... then, of one ony family there would be produced 111305, that is, One hundred and eleven thousand, three hundred and five Souls to begin the World .... ... such a number arising only from a simple production of only One yearly ...." I cannot work out how 111305 arises _ the fact that he spells it out makes it unlikely to be a misprint.

Ozanam. 1694. Prob. 8, 1696: 33-35; 1708: 29-32. Prob. 11, 1725: 68-75. Section II, 1778: 68-74; 1803: 70-76; 1814: ??NYS; 1840: 34-36. A discussion of geometric progression and a mention of 1, 2, 4, .... 1778 et seq also mention 1, 3, 9, ....

Ozanam. 1725. Prob. 11, questions 6 & 7, 1725: 79-82. Prob. 3, parts 1-3, 1778: 80-82; 1803: 82-84; 1814: 72-75; 1840: 38-39. Examples of population growth in Biblical and biological contexts. In 1725, he has ratios of 2, 50, 3, 4, 1000, The examples vary a bit between 1725 and 1778.

Walkingame. Tutor's Assistant. 1751. The section Geometrical Progression gives several problems with powers of 2 and the following less common types.

Prob. 5, 1777: p. 95; 1835: p. 103; 1860: p. 123. Find 1 + 4 + 16 + ... + 4¹¹ farthings.

Prob. 8, 1777: p. 96; 1835: p. 104; 1860: p. 123. Find 2 + 6 + 18 + ... + 2 x 3²¹. If these are pins, worth 100 to the farthing, what is the value?

Vyse. Tutor's Guide. 1771? The section Geometrical Progression, pp. 146-151 & Key pp. 190-192, gives several examples with doublings and triplings as well as examples with ratios of 3/2 and 10. There is a major error in the solution of prob. 7, to find 2 + 6 + 18 + ... + 2 x 3¹⁹.

Pike. Arithmetic. 1788. Pp. 237-239. Numerous fairly standard examples, mostly doubling, but with examples of powers of 3 and of 10 and the following.

Pp. 239-240, no. 8. One farthing placed at 6% compound interest in year 0 is worth what after 1784 years? And supposing a cubic inch of gold is worth £53 2s 8d, how much gold does this make? This is very close to 2¹⁵⁰ farthings and makes about 4 x 10¹⁴ solid gold spheres the size of the earth!

Eadon. Repository. 1794. P. 241, ex. 3. Doubling 20 times from a farthing.

John King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995. P. 100. 10 + 10² + ... + 10¹¹ grains of wheat, converted to bushels and value at 4s per bushel.

(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255 & 3:8 (Aug 1889) 351. (This is undoubtedly reprinted from Boy's Own Magazine 1 (1863).) Mathematical question 59. Seller of 12 acres asks 1 farthing for the first acre, 4 for the second acre, 16 for the third acre, .... Buyer offers £100 for the first acre, £150 for the second acre, £200 for the third acre, .... What is the difference in the prices asked and offered? Also entered in 7.AF.

Definizione.
Si dice progressione geometrica (o per quoziente) una successione di numeri tale che sia costante la quoziente fra un qualunque numero e il suo predecessore.

Esempio:

3, 6, 12, 24, ...

6/3 = 2
12/6 = 2
24/12 = 2
...

I termini successivi di una progressione si possono indicare così (lettere minuscole con indice):
a₁ , a₂ , a₃ , ..., a_n , ...

a₁ è il primo termine della progressione.

Il quoziente costane costante, che nell'esempio numerico è 2, si chiama ragione della progressione geometrica e si indica con q.

Possiamo quindi definire una progressione così:
a_n = a_n-1 ´ q

Formula per calcolare il termine n-esimo di una progressione geometrica conoscendo il primo termine e la ragione.
a_n= a₁ ´ q^n-1

Formula per calcolare la somma dei primi n termini di una progressione geometrica.
S = a₁(qⁿ - 1)/(q-1)

Formula per calcolare il prodotto dei primi n termini di una progressione geometrica.
P = ±Ö(a₁ ´ a_n)ⁿ