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# Title: Leibniz-Handschriften zur Mathematik LH 35, 12, 1
# Folio 329v, page id 00068194:0660
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Datis duobus terminis progressionis Hamoniae
inventie tertium santo, illi termini Ae+ A
a +h +b. Fy Invenire tertiuu harmonice
proportionalem Dati sunto Z. primus. Z. a.
secundus. Q- a + b. tertius; quaeritur b.
vel quarit atb, ralio differentiam
est a
Ratio termini primi
2 ad testium est
a  2.
vel B= 2- a +b
atb,
Dit
Multolicatudter lmin agintionus
per  fieta b2 = 4- a +b,
a 1+
ergo es
a
et b a
a +1 = 2a - 34
ba
(4
att) e
2+ a+1. = Z 2
a
ga-
ba-
1+ 1
ati = 2 ba ehqe 5 +1
ba
 a+1
2 se 11.
ba: ba.
p
sibbaa L
b= 2, 7a XB, seu 
2 +
t
A=
er
12 a-b
 b2- ZF atb.=
 sei 2 Z = Z -a- 10
e
seu Z + a = q a seu 7 +1
tia
dseu 
=
+1
Bt1 a 6 L f1
d- IseD
= 1B. 1 
Dato probler sic ergo rectius cocipretur
problema: dato termino primo seu
maximo proportionis harmonicae,
dataque  differentia maxima
(qua terminus primus differt a secundo)
et per consequens dato termino primo
ad et secundo; in venire differentiam
 secundam, termini secundi a tertio, seu
edium fertium. vel quod dem est
Ad continuare progressionem Harmonitam in
ainfinitum A
del
2a = ()
fatbo et
dC
1
32p a +b
ua
3
eanon t etf
e
e
a + e tf
+nC
e-5
1. L
2ca- 1C es +a
37fg.
odejus resolutio i
2aF  caeteras datas
fab se quaeritut
10
324 = dca = C 1d c- Qca. Dividantu
32
ber
na per C
ut 
Jam Zualet
Binao laradu
ax + Xt2bL(4 (1Z6 ab
agtharitea v.
2- ac
Ta
dic
c
e
b. ergo
+a
2. _E. Z. a = 5 in Z- (a - b - c
l.s 5+ e c C xd.
In prosie (o Harmonica, 2G ergo X - 1 ergo C=
Est ergo haec problematis solutio datis duobus
fier
7a-1 c H = 2a 5 v. -aa g2at ab
(4)
4 - aa eu 2+ 4. L-  gei 12+ a
26 a) 1, FI51A S(1) 3u(2
seu
d d. C = ASZ Ea  puo Z substoliat
24a 2a9a
2.1 = 27.1
maed (26 = 6+ 14+
 (a
e
terminis, invenires differentiam secundinuri item a
2. GEuo Zd = C hoe
a tertio seu tertiu. minimum. quad ita fiet:
A Terminus sependus, dividatu per 2.a = C Ergo ZC- a C = C. Et Z a - L seu
dibterenti Primgu dividatur per differens C =C: Item B a- b = s E Ergo
tiam plud actus addaber unrtate suma
a
aucto dividat terminum primum difet
2f- af - bf = C.. Et 2- a b +
rentia prima minutum Quotiens est differentia
Vatertia. Sic eego rectius concipiet problema.
L. 4. b f,d c 5
L.dib 3quf
la habemus has
tequatione