And (y3+u3) 72=x2z+xz2, and things that are equal to the same are equal; therefore 9z-z2.9=72, or 9z-z2=8, or z2−9z=−8; whence by completing the square, &c.z=8, æ= (9—z=) 1, y=(3 √/x2z=) 2, u=(3‚√/xz2=) 4.

5. Of four numbers in geometrical progression, the product of the two least is 8, and of the two greatest 128; what are the numbers?

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

Therefore (8 × 128=) 1024 =uy, or uy=32, and u=

extremes and means, y2=.

32

y

But (xy:: y: u, that is,)

:y y:

y

256

6. The sum of 3 numbers in geometrical progression is 14, and the greater extreme exceeds the less by 6; what are the numbers? Ans. 2, 4, and 8.

125. Def. Compound Interest is that which is paid for the use, not only of the principal or sum lent, but for both principal and interest, as the latter becomes due at the end of the year, half-year, quarter, or other stated time.

To investigate the rules of Compound Interest.

Let p=the principal, r= the rate per cent. t=the time, R= (1+r=) the amount of 1l. for a year, called the ratio of the rate per cent. a=the amount.

Then since 1 pound: is to its amount for any given time and rate :: so are any number of pounds: to their amount for the same time and rate; therefore as

1: R::

P : pR = the first,

pR : pR2= second,
pR : pR=third,
pR3 : pR = fourth,
PR-1 : pR= t

year's amount.

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, u have xz=y, (Art. 120. Note,) and (xyz=xz.y=) y3=100% whence y=10; also xz=(y2=) 100, and by the problem x+ 25: from the square of this equation subtract four times the pr ceding, and x2-2 xz+z2=225: extract the square root of th and x-z=15; add this to, and subtract it from, the equati x+z=25, and 2x=40, or x=20, also 2 z=10, or z=5; whe 5, 10, and 20, are the numbers.

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

Let n-2=the number of mean proportionals, then will the number of terms in the progression: also let r=the ratio, i

b

n

(theor. 3. Geom. Prog.) r=; and by logarithms, log. b—l

a

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

EXAMPLES.-1. To find two mean proportionals betwee

whence 12x7=84, the first mean, and 84x7=588, the

Whence we have THEOREM 1. pR-a, THEOr. 2.

α

R

of which follow immediately from the first; the fourth cannot be conveniently exhibited in numbers without the aid of logarithms.

By means of these four theorems, all questions of compound interest may be solved.

EXAMPLES.-1. What is the amount of 1250l. 10s. 6d. for 5 years, at 4 per cent. per annum, compound interest?

Here p (1250l. 10s. 6d.=) 1250,525, t=5, R=1.04. Then theor. 1.(pRt=) 1250.525 × 1.045=1250.525 × 1.2166 =1521.388715=1521l. 7s. 94d.=a.

2. What principal will amount to 2001. in 3 years, at 4 per cent. per annum ?

α 200

Here a 200, R=1.04, t=3, and theor. 2. (—)

200

1.124864

=

=177.7992=1771. 15s. 114d.=p.

R 1.043

=

3. At what rate per cent. per annum will 500l. amount to 578l. 16s. 3d. in 3 years?

theor.

Here p 500, a= =(578l. 16s. 3d.=) 578.8125, t=3; and,

1

P. 3. Art. 63.=) —×5.25=1.05=R: wherefore, (since R—1

=r,) we have R—1=.05=r, viz. 5 per cent. per annum.

4. In how many years will 2251. require to remain at interest, at 5 per cent. per annum, to amount to 2601. 9s. 3şd. ?

Here p=225, R=1.05, a=(260l. 9s. 34d.=) 260.465625:
log. a-log. p_log. 260.465625-log. 225
log. R

whence, theor. 4. (

log. 1.05

=3 years=t.

5. What sum will 500l. amount to in 3 years, at 5 per cent.

per annum? Ans. 5781. 16s. 3d.

6. What principal will amount to 15211. 7s. 94d. in 5 years, at 4 per cent. per annum? Ans. 1250l. 10s. 6d.

7. At what rate per cent. will 7217. amount to 1642l. 19s. 94d. in 21 years? Ans. 4 per cent.

8. In how many years will 7211. be at interest at 4 per cent. to amount to 1642l. 19s. 94d. Ans. 21 years.

If the interest be payable half-yearly, make t=the number of half-years, that is twice the number of years, and r=half the rate per cent. but if the interest be payable quarterly, let t=the number of quarter-years, viz. 4 times the number of years, and r=one-fourth of the rate per cent. and let R=r+1 in both cases, as before k.

126. To determine some of the most useful properties of

numbers.

Def. 1. One number is said to be a multiple of another, when the former contains the latter some number of times exactly, without remainder.

Thus 12 is a multiple of 1, 2, 3, 4, and 6.

COR. Hence every whole number is either unity, or a multiple of unity.

2. One number is said to be an aliquot part of another, when the former is contained some number of times exactly in the latter.

Thus 1, 2, 3, 4, and 6, are aliquot parts of 12, for 1 is TT, 2 is, 3 is 4, 4 is, and 6 is 4 of 12.

COR. Hence no number which is greater than half of another number, can be an aliquot part of the latter.

3. One number is said to measure another number, when it will divide the latter without remainder.

Thus each of the numbers 1, 2, 4, 5, 10, and 20, measures 20. 4. One number is said to be measured by another, when the latter will divide the former without remainder.

Thus 20 is measured by 1, 2, 4, 5, 10, and 20.

COR. Hence every aliquot part of a number measures that number, and every number is measured by each of its aliquot parts, and by itself.

k It was at first intended to investigate and apply every rule in arithmetic, but want of room obliges us to omit Equation of Payments, Loss and Gain, Barter, Fellowship, and Exchange; these will be easily understood from the doctrine of proportion, of which we have amply treated.