123. To shew how the 17th, 18th, 19th, and 20th theorems are derived.

Since Z=A+R.n-1 (th. 2.) therefore n-1=

Z-A
R

n= +1 (THEOR. 17.) and because R=L.s-a-L.s—z

(th. 4.) substitute this value for R in theor. 17. and n=

Z-A
L.s-a-L.s-z

+1 (THEOR. 18.) again, for Z in theor. 17. sub

Z-A

substitute its value from theor. 5. and n=

+1=)

R

EXAMPLES.-1. Given the ratio 2, the number of terms 6, and the last term 96, of a geometrical progression, to find the first term, and the sum of the terms?

Here r=2, n=6, z=96, whence (theor. 1.) a=

1

1.9822712

2. Given the ratio 2, the number of terms 6, and the sum of the terms 189, to find the first and last terms?

r-1.s

Here r=2, n=6, s=189, and (theor. 9.) a=

By Logarithms.

0.0000000 | Lr—1+R.n—1=

1.5051500

2.2764617

1.7993405 L.r−1+R.n—1+S= 3.7816117

2.2764617 | +S=

3. Given the first term 3, the ratio 2, and the last term 96, to find the number, and sum of the terms?

Here a=3, r=2, z=96, and (theor. 7.)

rz-a 2 × 96-3

-

1

L.189=2.2764617

-Lr-1=L.1=

0.0000000

S= 2.2764617

whence s=189.

....

R=0.3010300) 1.5051500(5

therefore n=5+1=6, theor 17.

4. Given the first term 4, the ratio 3, and the sum of the terms 484, to find the last term, and number of terms?

Here a=4, r=3, s=484, and (theor. 6.) z=

n the first term 2, last term 2048, and sum of the O, to find the ratio, and number of terms?

a=2, z=2048, s=2730, and (theor. 4.) r= =

2

2728

-L.s-z L.682=2.8337844

L.s-a-L.s-z=0.6020600

therefore .6020600)3.0103000(5
whence n=5+1=6, theor. 18.

Ans. a=

=en r=4, n=6, and s=2730, to find a and z. 048. _ven r=2, n=6, and z=96, to find a and s.. Ans. a=3,

ven the ratio 5, last term 12500, and sum of the terms to find the first term, and number of terms. Ans. a=4,

iven a=4, n=6, and z=12500, to find r and s. Answer

=15624.

Given r=3, n=4, and z=81, to find a and s.
Given r=6, n=5, and s=1555, to find a and z.
Given a=3, r=10, and n=20, to find s and z.

PROBLEMS IN GEOMETRICAL PROGRESSION. Of three numbers in geometrical progression, the difference first and second is 4, and of the second and third 12; red the numbers?

Let x, y, and z, be the numbers.

:y:

Then y—x=4, or x=y—4; z—y=12, or z=y+12. Wherefore since by the problem xy:: y: z, by substituting alues of x and z in this analogy, we shall have y —4 : y :: 12; wherefore, (by multiplying extremes and means,) y-4 12=) y2+8y—48=y2, or 8y=48; wherefore y=6, x=2,

18.

2. The product of three numbers in geometrical progression is 1000, and the sum of the first and last 25; required the numbers?

Let x, y, and z, be the numbers; then since xy::y:z, we have xz=y, (Art. 120. Note,) and (xyz=xz.y=) y3=1000, whence y=10; also xz=(y2=) 100, and by the problem x+z= 25: from the square of this equation subtract four times the preceding, and x2 — 2xz+z2=225: extract the square root of this, and x-z=15; add this to, and subtract it from, the equation x+z=25, and 2x=40, or x=20, also 2 z=10, or z=5; whence 5, 10, and 20, are the numbers.

3. To find any number of mean proportionals between two given numbers a and b.

Let n-2=the number of mean proportionals, then will n= the number of terms in the progression: also let r=the ratio, then (theor. 3. Geom. Prog.) r=

b; and by logarithms, log. b—log, a

a

÷n-1=log. r; whence r being found, if the less extreme be continually multiplied, or the greater divided, by r, the results will be the mean proportionals required.

EXAMPLES.-1. To find two mean proportionals between 12 and 4116.

Here a=12, b=4116, n=4, and r=

4116
12

=343+=)7;

whence 12×7=84, the first mean, and 84×7=588, the second

mean.

2. To find four mean proportionals between 2 and 486. Ans. 6, 18, 54, and 162.

3. To find five mean proportionals between 1 and 64.

4. There are four numbers in geometrical progression; the sum of the extremes is 9, and the sum of the cubes of the means 72; what are the numbers?

Let x, y, u, and z, be the numbers.

Then by the problem,

x+z=9, or x=9-z.

xy::u: z, or xz=uy, whence xz=(9—z.z=) 9z—2°.