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As in arithmetical progressions, all the questions which can be proposed for solution in geometric progressions reduce to 10, the solutions of which are deduced from

l=af>"-1 (1)

-£? w

from which it appears generally that as the number x increases, the difference of the logarithms of x and l~\-x diminishes. Also, since - is greater than the whole series, - being diminished by more than it is increased, we have

l(l+x)-U<-.

v x

If the base be 10, we have seen that M=0.4342. ..<-. Hence, in this case,

If x consist of 6ve places, its least value is 10000. Therefore the greatest value of

l(l4-x)Ix is less than ——=0.00005.

20000

Hence we may infer that the logarithms of every two consecutive whole numbers consisting of five places must agree in the first four decimal places at least. Now let

A = l{l+x)—i*=£±£.
A'=*(2+*)-/(l+x)=/?±|.

x l-f-j;
z(H-z) V x(2+x)>

But by [A], Art. 224,

'(1_r"i(2+x))=M \ x[2+x) ~ ±fi(S+x)^3x-'(2-(-x)3~*C' I

If X consist of five places, its least value is 10000, and, therefore, the greatest value of

A—A' is less than " , which, when reduced to a decimal, has no

20000X10002 200040000

significant figure within the first eight places. Hence, in tables which extend only to

seven places, we may assume that A—-A'=0, or A=A'.

Thus we infer that, under the circumstances which have been supposed, the logarithms

of numbers in arithmetical progression will themselves be in arithmetical progression.

Let now » and n-f-1 be two consecutive whole numbers, and an intermediate fraction. These may bo looked upon as three terms of an arithmetical progression, whose first

1 p term is n, whose common difference is -, whose (»-r-l)"' term is n-\—, and whose (<7-4-l)a'

1 9 term is n+1. By what has been already shown, the logarithms of the several terms of this scries will also be in arithmetical progression. Let 6 be their common difference. The (p-(-l),h term of this series will be

ln-\~p5,

which will be the logarithm of the (p+1)th term of the former scries;

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HARMONICAL PROGRESSION.

234. A series of quantities is called a harmonicak progression when, if any three consecutive terms be taken, the first is to the third as the difference of the first and second to the difference of the second and third.

Thus, if a, b, c, d.... be a series of quantities in harmonical progression, we shall have

a:c::a—6:6—c; b:d::bc:cd, &c.

235. The reciprocals of a series of terms in liarmonical progression are in arithmetical progression.

Let a, 6, c, d, e,f.... be a series in harmonical progression.
Then, by definition,
Alio, the laat term of the latter aeries, which will be

In+qi,

will be the logarithm of the laat term of the former series;

.•./(n+l)=/n+?<5, .-. I(n+l)—ln=qS.

B"»»TIB1. l(n+Z)-ln
Bat, alio,

(»+?)-»
'V _P

) —» 7

Hence the differences of the logarithms arc as the differences of the numbers.

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INTEREST AND ANNUITIES.

236. The solution of all questions connected with interest and annuities may be greatly facilitated by the employment of the algebraical formulae. In treating of this subject we may employ the following notation: Let; dollars denote the principal, r the interest of $1 for one year, t the interest of p dollars for t years.

s the amount of p dollars for t years at the rate of interest denoted by r.

t the number of years that; is put out at interest.

SIMPLE INTEREST.

Problem I.— To find the interest of a sum p for t years at the rate r.

Since the interest of one dollar for one year is r, the interest of; dollars for one year must be ; times as much, or;r; and for t years t times as much as for one year; consequently,

»=pfr (!)

Problem II.— To find the amount of a sum p laid out for t years at simple interest at the rate r.

The amount must evidently be equal to the principal, together with the interest upon that principal for the given time.

Hence s=p-\-ptr

=P(l + tr) (2)

EXAMPLE I.

Required the interest of $873.75 for 2i years at 4J per cent, per annum.
It will be convenient to reduce broken periods of time to decimals of a year.
By the formula (1) we have

i=ptr.

In the example before us,

p =8873.75 .

r =8.0475*

t =2| years =2.5 years.

.-. 1=873.75 X 2.5 X -0475 dollars. =8103.7578125. The amount of the above sum at the end of the given time will be s=d-\-dtr =8873.75+8103.757.

PRESENT VALUE AND DISCOUNT AT SIMPLE INTEREST.

The present value of any sum s due t years hence is the principal which in the time t will amount to s.

The discount upon any sum due t years hence is the difference between thai sum and its present value.

Problem III.— To find the present value of s dollars due t years hence, simple interest being calculated at the rate r.

By formula (2) we find the amount of a sump at the end of t years to be

s=p-\-ptr.

Consequently, p will represent the present value of the sum s due t years hence, and we shall have

for the expression required.

* r is the interest of 81 for one year. To find the value of r when interest is calculated

at the rate of 84} or 84.75 per cent, per annum, we have the following proportion:

8l00:81::84.75:r

.4.75 „

.\r=S =10.0475.

100

In like manner,

When the rate of interest per cent, is 87, then r=$0.07.

When the rate of interest per cent, is 6, then r= 0.06.

When the rate of interest per cent, is 5, then r= 0.05.

When the rate of interest per cent, is 4}, then r= 0.0475.

When the rate of interest per cent, is 4J, then r= 0.045.

When the rate of interest per cent, is 4», then r= 0.0425

When the rate of interest per cent, is 4, then r= 0.04.

When the rate of interest per cent, is 3}, then r= 0.0375.
dec. 4c. Itc.

Problem IV.— To find the discount on s dollars due t years hence, at the rale r, simple interest.

Since the discount on s is the difference between s and its present value, we shall havo

d=s—

i l + (r

str

=T+Tr<4>

EXAMPLE.

Required the discount on $100, due 3 months hence, interest being calculated at the rate of 5 per cent, per annum. Here s =$100

t = 3 months = .25 years.

r= s$ .05.

Here the present value of p is

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1.0125 =98.76543 dollars. But *=$100

^=$98.76543 .-. s—p or dis=$1.235.

ANNUITIES AT SIMPLE INTEREST.

Problem V.— To find the amount which must be paid at the end oft years, for the enjoyment of an annuity a, simple interest being allowed at the rate r.

At the end of the first year the annuity a will be due; at the end of the second year a second payment a will become due, together with ar the interest for one year upon the first payment; at the end of the third year a third payment a becomes due, together with 2ar the interest for one year upon the former two payments, and so on; the sum of all these will be the amount required.

Thusi

At the end of the first year, the sum due is a.
At the end of the second year, the sum due is a + ar.
At the end of the third year, the sum due is a-f-2ar.
At the end of the fourth year, the sum due is a-j-3ar.

ice. &c. &c.

At the end of the Ith year, the sum due is —l)ar.

Hence, adding these all together for the whole amount,

«=<<z-f-ar(l + 2-f3+ («—!■))•

Or, taking the expression for the sum of the arithmetical series, l-(-2-|-3 +

t(t—l)

s=la+ra.^-£- (5)

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