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

Z-A
R

and

+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=

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

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

—L.r"—1= 1.7993405 L.r−1 + R.n—1+S=3.7816117

....

A= ....0.4771212-L.r"—1=

whence a=3.

Z=.

whence z=96.

...1.5051500 2.2764617

1.7993405

1.9822712

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

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=

r-1.s+a

T

L.rz-a

L.189=2.2764617

—L.r-1=L.1=

0.0000000

S= 2.2764617

whence s=189.

5. Given the first term 2, last term 2048, and sum of the terms 2730, to find the ratio, and number of terms?

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

s-a

6. Given r=4, n=6, and s=2730, to find a and z. 2, z=2048.

Ans. a=

7. Given r=2, n=6, and z=96, to find a and s. Ans, a=3, $=189.

8. Given the ratio 5, last term 12500, and sum of the terms 15624, to find the first term, and number of terms. Ans. a=4, n=6.

9. Given a=4, n=6, and z=12500, to find r and s. r=5, s=15624.

Answer

10. Given r=3, n=4, and z=81, to find a and s. 11. Given r=6, n=5, and s=1555, to find a and z. 12. Given a=3, r=10, and n=20, to find s and z. 124. PROBLEMS IN GEOMETRICAL PROGRESSION. 1. Of three numbers in geometrical progression, the difference of the first and second is 4, and of the second and third 12; required the numbers?

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

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 the values of x and z in this analogy, we shall have y −4: y :: y : y+12; wherefore, (by multiplying extremes and means,) y—4 •y+12=) y2+8y-48=y2, or 8y=48; wherefore y=6, x=2,' z=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 x2-2 xz+z2=225: extract the square root of this, and `x-2=15; add this to, and subtract it from, the equation x+z=25, and 2 x=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=2; 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

=3431+=)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.

x: y::u:z, or xz=uy, whence xz=(9—z.z=) 9 z—z2.