than the difference between the first and second; since therefore magnitudes continually increasing are added to the first, we will at length come to a magnitude greater than any assigned.

It is necessary to prove that the increments are continually increasing, for if they were decreasing the sum might never exceed a certain quantity.

Note 2. But if the given ratio be of greater inequality, we may at length come to a magnitude less than any assigned.

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It is thus demonstrated. Suppose the given terms of the series are assumed on a right line from one and the same point in it, and that they are as 16: 8, then assume any other right line, ever so small, suppose, then as the second term of the given ratio is to the first, i. e. as 8: 16, so let this assumed line , be to another part assumed in the same right line, and from the same point in it, i. e. for :::8:16; let this series of be continued till a magnitude is found greater than (16) the first term of the given series (by preceding note), i. e. let it be continued through the terms,, 1, 2, 4, 8, 16, 32; then let the given series of 16: 8 be continued down through as many terms as there are up from to 32, i. e. let it be continued through the terms 16, 8, 4, 2, 1, 1, 1, 1, and let the least term be; this is less than the assigned quantity, and so on we may continue the series through ever so many terms, without ever coming to an end, and the sum of all those terms, though always coming nearer to the first (16), can never be equal to it, even though continued to ; and .. they can never exceed the the assigned quantity, though they will approach nearer to it than any assignable difference.

Proof. Since there are two series of magnitudes continually proportional and equal in number, viz. 16, 8, 4, 2, 1,,,, and 32, 16, 8, 4, 2, 1, 4, 1, ex æquo, the first term is to the last in the first series as the first is to the last in the second series, i. e. 16::: 32: ; but the first term of the first series is less than the first term of the last,.. the last term of the first series is less than that of the last; but this last term of the last series is presumed a very small quantity, yet we have found a quantity

smaller than this; in like manner a quantity can be found less than any assumed quantity, be it ever so small.

Tacquet gives from Gregorius a S. Vincentio, a method of finding a series of right lines in any given ratio of greater inequality, and of exhibiting the sum of the series continued through an infinite number of terms; assuming those premises proved in the two last notes.

On any right line assume a part to the antecedent, or to the first term of the series; from the extremities of this assumed antecedent raise perpendiculars respectively = to it, and the consequent. Connect the extremities of those perpendiculars, and produce the connecting line to meet the line on which the part is assumed. That line will be the base of a right angled A; assume on it from the point in which the second perpendicular meets it, another part to this second perpendicular (or to the given consequent), and from the extremity of this perpendicular raise a perpendicular to meet the connecting line (or the hypothenuse); that perpendicular will be the third term of the series, and so on the series may be continued through any number of terms, by assuming on the base from the extremity of the last perpendicular a segment to it, and again raising from the extremity of this segment another perpendicular.

PART I. Those perpendiculars, or the segments into which the base is divided, are in continued proportion; for the whole base the difference between it and the first perpendicular i. e. the second base :: the first perpendicular the second (by similar As), or as the first assumed part the second, and .. permutando, the whole base : the first assumed part as the second base the second assumed part, and convertendo, the whole base: the second base the second base: the third base; but the first base the second base:: the first perpedicular : the second, and the second base the third base: the second perpendicular the third perpendicular; and thence the first perpendicular: the second the second the third. In like manner it can be shewn that : the other perperdiculars are proportional.

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PART II. Since the first perpendicular: the second :: the first base the second base; and as the first perpendicular is less than the first base, the second perpendicular will be less than the second base; in like manner the third,

fourth, fifth, &c. perpendiculars will be respectively less than the third, fourth, fifth, &c. bases; but the first perpendicular is to the first assumed part, the second perpendicular to the second assumed part, and the third to the third, &c. therefore, the whole series of proportionals is not greater than the first base; and since the bases themselves are continually proportional, the last term will be less than any assigned magnitude, and therefore the sum of all the assumed parts, or of the entire series of proportionals, if the number of terms be infinite, is not less than the entire base.

The difference between the first and second terms, the first term, and the sum of the entire series, are continually proportional.

For this difference: to a line to the first term: : the first perpendicular: the sum; but the first perpendicular is also to the first term, .. &c.

PROP. XII. PROB.

To find a fourth proportional to three given right lines.

Form a rectilineal and on one of its legs from the vertex, assume a part to the first antecedent, and in continuum with this assume another part = to the given consequent; on the other leg assume from the vertex a part to the second antecedent, and connect its extremity with the extremity of the first assumed antecedent; from the extremity of the assumed consequent draw a line parallel to this connecting line, the intercept (between those parallels,) of the other leg shall be the fourth proportional.

For (by Prop. 2. 6) the first assumed antecedent: assumed consequent second assumed antecedent: intercept, .. &c.

PROP. XIII. PROB.

To find a mean proportional between two given right lines.

Draw any right line, and on it take one after another parts respectively to the given extremes. Bisect the sum of those parts, and with the point of bisection as

centre and half sum as radius describe on the sum a semicircle; from the common extremity of the parts erect a perpendicular to meet the circumference, it will be the mean proportional required.

For draw chords from the extremity of this perpendicular at the circumference, to the extremities of the diameter; the contained by those chords will be a right (Prop. 31. 3), and the perpendicular from this right on the hypothenuse is a mean proportional between the seg. ments of the hypothenuse (Cor. 8. 6).

Cor. 1. In like manner mean proportionals can be found between the given lines and this mean, from whence a series of five right lines continually proportional would arise; and again, mean proportionals being found between the adjacent terms in this last series, a series of nine proportionals arises, and so on. But the number of proportionals in any series will be always one less than double the number of terms in the preceding series.

Cor. 2. Given of three proportionals, the sum of the extremes, and the mean, the extremes themselves can be found.

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On the given sum describe a semicircle, from either extremity of the diameter raise a perpendicular to the given mean, through the extremity of this draw a line parallel to the diameter to meet the circumference, and from the point in which it meets the circumference let fall a perpendicular on the given sum; this perpendicular will divide the given sum into the required segments. For it is a mean proportional between the segments (by prop.) and to the perpendicular raised from the extremity, i. e. to the given mean.

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Note. This proposition, with the 14th of second, proves, that the of the mean is to the rectangle under the extremes. For if we are given two numbers, suppose 4 and 9, we can find a mean proportional between them, by extracting the ✔ of their product, i. e. 36 = 6; then 6 is a mean proportional between 4 and 9. Note 2. In Cor. 2. the par. to the diameter will meet the periphery, for it is evident that the given mean cannot be greater than half the sum of the extremes.

Note 3.

If we have the sum of two right lines and the rectangle under them, we can find the lines; for the✔

2

of the rectang. or the side of a = the rectange, is =, the mean proportional between them. Thus, in numbers; let 30 be the sum of the numbers and 9 the mean; then the 2 of half the sum the of the mean the 2 of

2

intermediate part i. e. 225—81 = 144 and

144 = 12, which taken from half the sum, leaves 3 for one extreme, and added to half the sum makes 27 the other.

No method, purely geometrical, has been as yet discovered for finding two mean proportionals between two given right lines; however the following methods have been suggested by Plato, Philo of Byzantium, and Des Cartes, which are partly geometrical, and partly me.chanical.

Plato's Method.

For this method we must have an instrument composed of a mechanick's square, (i. e. two rulers fixed together at right s), and another ruler moveable along one side of this square so that it will be always at right As to it; then placing the given extremes at rights to one another, produce them beyond the vertex; and so apply the instru ment that the side of the square which has not the moveable ruler to it, may always pass through the extremity of the first extreme, and that the vertex of its < may lie on the other produced extreme; then the whole apparatus must be so arranged, that the moveable ruler will pass through the extremity of the last extreme, and that the first extreme produced will pass through the vertex of the <which this ruler makes with the side of the square, along which it moves. Then it is evident that those produced parts intercepted between the <formed by the extremes and the vertices of the instrument are the required means.

Philo's Method.

Let the given extremes be placed at rights to one another, complete the rectangle, and about it circumscribe a circle; then produce beyond the extremities of the extremes the sides opposite to them, and so apply a ruler to the vertex of the right formed by the extremes, that the portions of it between the produced parts