then will the two second and the two fourth terms form a new

proportion, and we shall have b: d::f: g.

For, from the first, or a b::c: d, is derived,

Again, if the two proportions be such that the mean terms are the same in both, then will the first terms be in an inverse proportion to the fourth terms. For instance, if

a: bc d; and f: b::c: g;

then will af::g: d.

For from the first, ad=bc; and from the second, fg be. .. ad fg. And consequently, a g::f: d,

or, a f::g: d.

348. Two proportions being given, a new one may always be produced, by multiplying the first term of the one by the first term of the other; the second of the one by the second of the other, and so on of the others. For example, from the proportions, a: b::c: d, and e: f::g: h, we obtain,

ae: bfcg: dh.

For the first proportion gives ad=bc, and the second gives eh=fg; we shall thus have,

adeh bcfg.

Now adeh is the product of the extremes, and bcfg is the product of the means of the new proportion; and as these two products are equal, we know that the proportion is true. Thus, if 6 4:15: 10, and 9: 12::15: 20,

then will, 6x9: 4 x 12::15 × 15: 10 x 20,

or, 54: 48:225: 200.

349. If

349. If three quantities be proportional, the product of the extremes is equal to the square of the mean; for if a: b::b: c, then (Art. 340.) ac-b2. From this it is evident that a mean proportional between any two numbers may be found by taking the square root of their product. For example, if it were required to find a mean proportional between 9 and 4, assume a for the mean required: then 9:x::x: 4; and by multiplying means and extremes, x=36, and r=6, the mean required.

350. If four quantities be proportional, then will any powers, or roots, of those quantities be likewise proportional: for if a: b::c: d, then will and therefore; and b-ď

a C

a2 c2

a2: b2:: c2: d2; and in general an: b"::c": d"; where n, being either integral or fractional, represents any power or root whatever, of the quantities a, b, c, d.

EXAMPLES.

1. If x-y: x+y in the triplicate ratio of a: b, and a : b:: 32, show that 19x=35y.

2. Divide the number 50 into two parts, which shall be to each other in the duplicate ratio of 3:1; and find a mean proportional between them.

Let the lesser part.

Then, since the two parts are to each other ::9: 1, assume 9x for the greater.

.. 9x+x, or 10x=50, and x=5, the lesser part;
and 9x=45, the greater.

Hence (Art. 349.) √45 × 5225 15 is the mean proportional required.

3. If a+bc::d+ef, and ac::df, show that b:c::e:f.

y2

Since a+bc::d+e:f
and a c::df.
..a+bad+e: d;

..ba: ed.
But a cdf.

.bce:f

4. If 22—Y°=ab, show that x+y: 4a::b: x—y.

5. There are two numbers, whose product is 45, and the difference of their squares: the squares of their difference:: 72. What are the numbers?

ANS. 9 and 5.

CHAP. XLIV.

OF GEOMETRICAL PROGRESSION.

351. A SERIES of numbers, which are continually increased or diminished by the multiplication or division of them by the same quantity, is termed a series in geometrical progression, because each term is to the succeeding one in the same geometrical ratio. The number which expresses how many times each number is greater or less than the preceding one is called the exponent, or ratio of the progression. Thus if the first term be 1, and the ratio 2, the progression would be as follows:

Terms 1 2 3 4 5 6 7 8

Prog. 1, 2, 4, 8, 16, 32, 64, 128, &c.

where the numbers 1, 2, 3, &c. denote the place which each term holds in the progression.

352. Now let us suppose generally the first term to be a, and the ratio r; we shall then have the following progression: Terms 1 2 3 4 5 6

Prog. a, ar, ar2, ar3, art, ar3, &c.

where we shall observe this comparison, viz. that the index of the ratio r is always less by unity than the term which it occupies in the progression; so that, if the progression were continued, the 7th term would be ar6; the 8th would be ar7; and the nth term would therefore be a-1.

353. Now if r be greater than unity, it is evident that the terms will continually increase, and on the contrary if it be less than 1, or a fraction, the terms will continually diminish. Thus if a 1 and r, the progression would be,

1,

1 1 1 1 1 1

2' 4' 8' 16' 32' 64'
&c.

354. We

354. We have therefore to consider,

1 The first term, or a.

2. The ratio, or r.

3. The number of terms, or n.
4. The last term, or l.

355. One of the principal questions which occurs on this subject, is to find the sum of all the terms of a geometrical progression. In order fully to explain the method by which this is accomplished, let us begin by taking the following progression, continued to ten terms, viz.

1, 2, 4, 8, 16, 32, 64, 128, 256, 512.

the sum of which being represented by S, we shall have the following equation,

S=1+2+4+8+16+32 +64 +128+256 +512.

Multiplying each side by the ratio 2, we have,

2 S=2+4+8 +16+32 +64 + 128 +256 +512+ 1024 Subtracting the first equation from the second there remains, S=1024-1=1023,

which we therefore find to be the sum of the progression required.

356. Supposing therefore generally that the first term is a, the common ratio r, and the number of terms n, we shall then have,

S=a+ar+ar2+ar3+ &c.......ar"-2+ar-1...

Multiplying this equation by the ratio r, we have,

r Sar, +are+ar3+ &c. ...... ar+ar....

Subtracting the first equation from the second,

rS-S-ar"-a, or (r-1). S-ar"-a,