Charles Davies.

First lessons in algebra, embracing the elements of the science online

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1 . 4 . 7 . 10 . 13 . 16 . 19 . . .,
we have J=l-f 49x3 = 148.

13. To find the 60th term of the progression

1 . 5 . 9 . 13 . 17 . 21 . 25 . . .,
we have Z= 1+59x4=237.

\ 139. If the progression were a decreasing one, we
should have

l=a— (n — l)r.

Hence, to find the last term of a decreasing progression,
we have the following

RULE.

I. Multiply the common difference by one less than the num-
ber of terms.

II. Subtract the product from the first term : the remainder
will be the last term.



Quest. — 139. Give the rule for finding the last terra of a series*
when the progression is decreasing.



ARITHMETICAL PROGRESSION. 221



EXAMPLES.

1. The first term of a decreasing progression is 60, the
number of terms 20, and the common difference 3 : what
is tne last term 1

l~a — (n—l)r gives 1=60— (20 — 1)3 = 60 — 57=3.

2. The first term is 90, the common difference 4, and
the number of terms 1 5 : what is the last term ? Ans. 34.

3. The first term is 100, the number of terms 40, and the
common difference 2 : what is the last term 1 Ans. 22.

4. The first term is 80, the number of terms 10, and the
common difference 4 : what is the last term ? Ans. 44.

5. The first term is 600, the number of terms 100, and
the common difference 5 : what is the last term ?

Ans. 105.

6. The last term is 8,00, the number of terms 200, and
the common difference 2 : what is the last term ?

Ans. 402.

140. A progression by differences being given, it is
proposed to prove that, the sum of any two terms, taken at
equal distances from the two extremes, is equal to the sum of
the two extremes.

That is, if we have the progression

2 . 4 . 6 . 8 . 10 . 12,

we wish to prove that

4+10 or 6+8

is equal to the sum of the two extremes 2 and 12.

19*



222 FIRST LESSONS IN ALGEBRA.

Let a.b.c.d.e.f....i.k.l be the pro-
posed progression, and n the number of terms.

We will first observe that, if x denotes a term which has
p terms before it, and y a term which has p terms after it,
we have, from what has been said,

x = a-j-pxr,

and y=l—px r ;

whence, by addition, x-\-y—a-\-l.

"Which demonstrates the proposition.

Referring this proof to the previous example, if we sup-
pose, in the first place, x to denote the second term 4, then
y will denote the term 10, next to the last. If x denotes
the 3rd term 6, then y will denote 8, the third term from
the last.

Having proved the first part of the proposition, write the
progression below itself, but in an inverse order, viz :

a.b.c.a.e.f...i.k.l.

I . k . i c . b . a.

Calling S the sum of the terms of the first progression,
2S will be the sum of the terms in both progressions, and
we shall have

2S={a+l) + (b+k) + {c+t) . . . +(i +c ) + (k+b)+(l+a).

Now, since all the parts a-\-l, b + k, c-\-i . . . are equal
to each other, and their number equal to n,

2S={a+l)n, or S=(^p-)n.



ARITHMETICAL PROGRESSION. 223

Hence, for finding the sum of an arithmetical series, we
have the following



RULE.

I. Add the two extremes together, and take half their sum.

II. Multiply the half-sum by the number of terms ; the
product will be the sum of the series.

EXAMPLES.

1. The extremes are 2 and 16, and the number of terms
8 : what is the sum of the series ?

S=(^±i)xn, gives S=^±^X8 = 72.

2. The extremes are 3 and 27, and the number of terms
12 : what is the sum of the series 1 Ans. 180.

3. The extremes are 4 and 20, and the number of terms
10 : what is the sum of the series 1 Ans. 120.

4. The extremes are 100 and 200, and the number of
terms 80 : what is the sum of the series ? Ans. 12000.

5. The extremes are 500 and 60, and the number of terms
20 : what is the sum of the series ? Ans. 5600.

6. The extremes are 800 and 1200, and the number of
terms 50 : what is the sum of the series ? Ans. 50000.



Quest. — 140. In every progression, what is the sum of the two ex-
tremes equal to 1 What is the rule for finding the sum of an arithmeti
cal series 1



224 FIRST LESSONS IN ALGEBRA.

> v 1 4 1 . In arithmetical proportion there are five numbers
to be considered : —

1st. The first term, a.

2nd. The common difference, r.

3rd. The number of terms, n.

4th. The last term, I.

5th. The sum, S.

The formulas

l=a+{n-\)r and S=(^tL\ X n

contain five quantities, a, r, n, Z, and S, and consequently
give rise to the following general problem, viz : Any three
of these jive quantities being given, to determine the other
two.

We already know the value of S in terms of a, n, and r.
From the formula

l=a+(n—l)r,
we find a=l— (n— l)r.

That is : The first term of an increasing arithmetical pro-
gression is equal to the last term, minus the product of the
common difference by the number of terms less one.
From the same formula, we also find

I — a

r =—v

That is : In any arithmetical progression, the common differ-
ence is equal to the difference between the two extremes divided
by the number of terms less one.

Quest. — 141. How many numbers are considered in arithmetical
proportion 1 What are they 1 In every arithmetical progression, what
is the common difference equal to 1



ARITHMETICAL PROGRESSION. 225

The last term is 16, the first term 4, and the number of
terms 5 : what is the common difference ?

The formula r=

n—\

16—4 m

gives r=z =3.

4

2. The last term is 22, the first term 4, and the number
of terms 10 : what is the common difference ? Ans. 2.

142. The last principle affords a solution to the follow-
ing question :

To find a number m of arithmetical means between two
given numbers a and b.

To resolve this question, it is first necessary to find the
common difference. Now, we may regard a as the first
term of an arithmetical progression, b as the last term, and
the required means as intermediate terms. The number of
terms of this progression will be expressed by m+2.

Now, by substituting in the above formula, b for /, and
7tt-f-2 for n, it becomes

b — a b — a



m + 2 — 1 m-j-l



that is, the common difference of the required progression is
obtained by dividing the difference between the given num-
bers a and b, by one more than the required number of
means.



Quest. — 142. How do you find any number of arithmetical means
between two given numbers 1



226 FIRST LESSONS IN ALGEBRA.

Having obtained the common difference, form the second
term of the progression, or the first arithmetical mean, by

adding r, or , to the first term #. The second mean

is obtained by augmenting the first by r, &c.

1. Find three arithmetical means between the extremes
2 and 18.

The formula r=

18-2 „
gives r= — - — =4 ;

hence, the progression is

2 . 6 . 10 . 14 . 18.

2. Find twelve arithmetical means between 12 and 77.
b-a



The formula



gives



rn-t-1

77-12



13
Hence the progression is

12 . 17 . 22 . 27 .... 77

143. Remark. If the same number of arithmetical
means are inserted between all the terms, taken two and
two, these terms, and the arithmetical means united, will
form but one and the same progression.

For, let a.b.c.d.e.f... be the proposed
progression, and m the number of means to be inserted
between a and b, b and c, c and d . . .



ARITHMETICAL PROGRESSION. 227

From what has just been said, the common difference of
each partial progression will be expressed by

b — a c — b d—e



m+l ' m+l 7 m+1

which are equal to each other, since a, b, c . . . are in
progression : therefore, the common difference is the same
in each of the partial progressions ; and since the last term
of the first, forms the first term of -the second, &c, we may
conclude that all of these partial progressions form a single
progression.



EXAMPLES.

1 . Find the sum of the first fifty terms of the progression
2 . 9 . 16 . 23 . . .

For the 50th term we have

/z=2 + 49x7.-=345.



50
Hence, S = (2 + 345) X— =347x25 = 8675.



2. Find the 100th term of the series 2 . 9 . 16 . 23 . . .

Ans. 695.

3. Find the sum of 100 terms of the series 1.3.5.
7.9 ... Ans. 10000.

4. The greatest term is 70, the common difference 3, and
the number of terms 21 : what is the least term and the
sum of the series 1

Ans. Least term 10 ; sum of series 840.



228 FIRST LESSONS IN ALGEBRA.

5. The first term is 4, the common difference 8, and the

number of terms 8 : what is the last term, and the sum of

the series ?

. C Last term 60.

Ans. < -

( Sum := 256.

6. The first term is 2, the last term 20, and the number
of terms 10 : what is the common difference ?

Ans. 2.

7. Insert four means between the two numbers 4 and 19 :
what is the series 1

Ans. 4 . 7 . 10 : 13 . 16 . 19.



8. The first term of a decreasing arithmetical progres-
sion is 10, the common difference — , and the number of

o

terms 21 : required the sum of the series.

Ans. 140.

9. In a progression by differences, having given the
common difference 6, the last term 185, and the sum of the
terms 2945 : find the first term, and the number of terms.

Ans. First term =5 ; number of terms 31.

10. Find nine arithmetical means between each antece-
dent and consequent of the progression 2.5.8.11.14 ...

Ans. Common dif, or r=0,3.

1 1 . Find the number of men contained in a triangular bat-
talion, the first rank containing one man, the second 2, the
third 3, and so on to the »*, which contains n. In other
words, find the expression for the sum of the natural num-
bers 1, 2, 3 . . ., from 1 to n inclusively.

Ans. S=&±9.



GEOMETRICAL PROPORTION. 229

9. Find the sum of the n first terms of the progression
of uneven numbers 1, 3, 5, 7, 9 . . . Arts* S = n 2 .

10. One hundred stones being placed on the ground in a
straight line, at the distance of 2 yards from each other,
how far will a person travel who shall bring them one by
one to a basket, placed at 2 yards from the first stone ?

Arts. 11 miles, 840 yards.



Geometrical Proportion and Progression.

144. Ratio is the quotient arising from dividing one
quantity by another quantity of the same kind. Thus, if
the numbers 3 and 6 have the same unit, the ratio of 3 to 6
will be expressed by

£*

And in general, if A and B represent quantities of the same
kind, the ratio of A to B will be expressed by

B_
A'

145. If there be four numbers

2, 4, 8, 16,

having such values that the second divided by the first is
equal to the fourth divided by the third, the numbers are



Quest.— 144. What is ratio 1 What is the ratio of 3 to 6 1 Of 4
to 121

20



230 FIRST LESSONS IN ALGEBRA.

said to be in proportion. And in general, if there be four
quantities, A, B, C, and D, having such values that

B__B_
A~C

then A is said to have the same ratio to B that C has to D ;
or, the ratio of A to B is equal to the ratio of C to D.
When four quantities have this relation to each other, they
are said to be in proportion. Hence, proportion is an equality
of ratios.

To express that the ratio of A to B is equal to the ratio
of C to D, we write the quantities thus :

A : B : : C : D ;

and read, A is to B as C to D.

The quantities which are compared together are called
the terms of the proportion. The first and last terms are
called the two extremes, and the second and third terms, the
two means. Thus, A and D are the extremes, and B and
C the means.

140. Of four proportional quantities, the first and third
are called the antecedents, and the second and fourth the
consequents ; and the last is said to be a fourth proportional
to the other three taken in order. Thus, in the last pro-
portion A and C are the antecedents, and B and D the
consequents.



Quest. — 145. What is proportion 1 How do you express that four
numbers are in proportion] What are the numbers called'? What are
the first and fourth called] WTiat the second and third? — 146. In four
proportional quantities, what are the first and third called \ Wliat the
second and fourth ]



GEOMETRICAL PROPORTION. 231

147. Three quantities are in proportion when the first
has the same ratio to the second that the second has to the
third ; and then the middle term is said to be a mean pro-
portional between the other two. For example,

3 : 6 : : 6 : 12;

and 6 is a mean proportional between 3 and 12.

148. Quantities are said to be in proportion by inver-
sion, or inversely, when the consequents are made the ante-
cedents and the antecedents the consequents.

Thus, if we have the proportion

3 : 6 : : 8 : 16,
the inverse proportion would be

6 : 3 : : 16 : 8.

1 49. Quantities are said to be in proportion by alterna-
tion, or alternately, when antecedent is compared with ante-
cedent and consequent with consequent.

Thus, if we have the proportion

3 : 6 : : 8 : 16,
the alternate proportion would be

3 : 8 : : 6 : 16.



Quest. — 147. When are three quantities proportional? What is the
middle one called 1 — 148. When are quantities said to be in proportion
by inversion, or inversely 1 — 149. When are quantities in proportion by
aUernation 1



232 FIRST LESSONS IN ALGEBRA.

150. Quantities are said to be in proportion by compo-
sition, when the sum of the antecedent and consequent is
compared either with antecedent or consequent.

Thus,, if we have the proportion

2 : 4 : : 8 : 16,
the proportion by composition would be

2 + 4 : 4 : : 8+16 : 16;
that is, 6:4:: 24 : 16.

151. Quantities are said to be in proportion by division,
when the difference of the antecedent and consequent is
compared either with antecedent or consequent.

Thus, if we have the proportion

3 : 9 ; : 12 : 36,
the proportion by division will be

9-3 : 9 : : 36-12 : 36;
that is, 6 : 9 : : 24 : 36.

152. Equi-multiples of two or more quantities are the
products which arise from multiplying the quantities by the
same number.

Thus, if we have any two numbers, as 6 and 5, and mul-
tiply them both by any number, as 9, the equi-multiples will
be 54 and 45 ; for

6x9=54, and 5x9=45.



Quest. — 150. When are quantities in proportion by composition?
— 151. When are quantities in proportion by division 1 — 152. What
are equi-multiples of two or more quantities 1



GEOMETRICAL PROPORTION. 233

Also, mxA and mx B are equi-multiples of A and B, the
common multiplier being m.

153. Two quantities, A and B, are said to be recipro-
cally proportional, or inversely proportional, when one in-
creases in the same ratio as the other diminishes. When
this relation exists, either of them is equal to a constant
quantity divided by the other.

Thus, if we had any two numbers, as 2 and 4, so related
to each other that if we divided one by any number we must
multiply the other by the same number, one would increase
just as fast as the other would diminish, and their product
would be constant.

154. If we have the proportion

A : B : : C : D,

7? D
we have —=z-—, (Art. 145);

and by clearing the equation of fractions, we have

BC=AD.

That is, Of four proportional quantities, the product of the
two extremes is equal to the product of the two means.

This general principle is apparent in the proportion be-
tween the numbers

2 : 10 : : 12 : 60,
which gives 2x60 = 10x12 = 120.

Quest. — 153. When are two quantities said to be reciprocally pro-
portional 1 — 154. If four quantities are proportional, what is the product
of the two means equal to 1

20*



234 FIRST LESSONS IN ALGEBRA.

155. If four quantities, A, B, C, D, are so related to
each other that

AxD=:BxC 9

P 7)

we shall also have -——-—;

A G

and hence, A : B : : C : D.

That is : If the product of two quantities is equal to the pro-
duct of two other quantities, two of them may he made the
extremes, and the other two the means of a proportion.
Thus, if we have

2x8 = 4x4,

we also have

2 : 4 : : 4 : 8.

1 56. If we have three proportional quantities
A : B : : B : C,

I B C

we have —.=-=7 ;

A B
hence, B 2 =AC.

That is : The square of the middle term is equal to the product
of the two extremes.

Thus, if we have the proportion

3 : 6 : : 6 : 12,
we shall also have

6x6=:6 2 =:3xl2 = 36.

Quest. — 155. If the product of two quantities is equal to the product
of two other quantities, may the four be placed in a proportion 1 How 1
— 156. If three quantities are proportional, what is the product of the
extremes equal to I



GEOMETRICAL PROPORTION. 235

157. If we have

A : B : : C : D, and consequently — =— ,

C

multiply both members of the last equation by —, we

then obtain,

CD
A~B'

and hence, A : C : : B : D.

That is : If four quantities are proportional, they will be in
proportion by alternation.

Let us take, as an example,

10 : 15 : : 20 : 30.
We shall have, by alternating the terms,

10 : 20 : : 15 : 30.

158. If we have

A : B : : C : D and A : B : : E : F,
we shall also have

B D . B F
A = ~C and -A=E ;

D F
hence, — =— and C : D : : E : F.

That is : If there are two sets of proportions having an



Quest. — 157. If four quantities are proportional, will they be in pro-
portion by alternation !



236 FIRST LESSONS IN ALGEBRA.

antecedent and consequent in the one equal to an antecedent
and consequent of the other, the remaining terms will be pro-
portional.

If we have the two proportions

2 : 6 : : 8 : 2*4 and 2 : 6 : : 10 : 30,

we shall also have

8 : 24 : : 10 : 30.

159. If we have

B D

A : B : : C : D, and consequently —=:—,

we have, by dividing 1 by each member of the equation,

A C

— =— , and consequently B : A : : D : C.
n D

That is : Four proportional quantities will be in proportion,
when taken inversely.

To give an example in numbers, take the proportion

7 : 14 : : 8 : 16;

then, the inverse proportion will be

14 : 7 : : 16 : 8,

in which the ratio is one-half.

160. The proportion

A : B :: C : D gives AxD=BxC.



Quest. — 158. If you have two sets of proportions having an ante-
cedent and consequent in each, equal; what will follow 1 — 159. If four
quantities are in proportion, will they be in proportion when taken in-
versely 1



GEOMETRICAL PROPORTION. 237

To each member of the last equation add BxD. We
shall then have

(A+B)xD=(C+D)xB;

and by separating the factors, we obtain

A+B : B : : C+D : D.

If, instead of adding, we subtract BxD from both mem-
bers, we have

(A-B)xD=(C~D)xB;
which gives

A-B : B : : C-D : D.

That is : If four quantities are proportional, they will be in
proportion by composition or division.

Thus, if we have the proportion

9 : 27 : : 16 : 48,

we shall have, by composition,

9+27 : 27 : : 16 + 48 : 48 ;

that is, 36 : 27 : : 64 : 48,

in which the ratio is three-fourths.

The proportion gives us, by division,

27-9 : 27 : : 48-16 : 48;

that is, 18 : 27 : : 32 : 48,

in which the ratio is one and one -half.



Quest. — 160. If four quantities are in proportion, will they be in pro-
portion by composition 1 Will they be in proportion by division 1 "What
is the difference between composition and division 1






238 FIRST LESSONS IN ALGEBRA.

1 6 1 . If we have

b i>

and multiply the numerator and denominator of the first
member by any number m, we obtain

-=— - and mA : mB : : C : D.



mA C

That is : Equal multiples of two quantities have the same
ratio as the quantities themselves.

For example, if we have the proportion

5 : 10 : : 12 : 24,

and multiply the first antecedent and consequent by 6, we
have

30 : 60 : : 12 : 24,

in which the ratio is still 2.

162. The proportions

A : B : : C : D and A : B : : E : F,

give AxD=BxC and AxF^BxE;

adding and subtracting these equations, we obtain

A{D±F) = B(C±E), or A : B : : C±E : D+F.

That is : If C and D, the antecedent and consequent, be aug-
mented or diminished by quantities E and F, which have the
same ratio as C to D, the resulting quantities will also have
the same ratio.



Quest. — 161. Have equal multiples of two quantities the same ratio
as the quantities? — 162. Suppose the antecedent and consequent be
augmented or diminished by quantities having the same ratio ?



GEOMETRICAL PROPORTION. 239

Let us take, as an example, the proportion

9 : 18 : : 20 : 40,

in which the ratio is 2.

If we augment the antecedent and consequent by 15 and
30, which have the same ratio, we shall have

9+15 : 18 + 30 : : 20 : 40 ;

that is, 24 : 48 : : 20 : 40,

in which the ratio is still 2.

If we diminish the second antecedent and consequent by
the same numbers, we have

9 : 18 : : 20-15 : 40-30;

that is, 9 : 18 : : 5 : 10,

in which the ratio is still 2.

163. If we have several proportions

A : B : : C : D, which gives AxD — BxC,
A : B : : E : F, „ „ AxF=BxE,
A : B : : G : H, „ „ AxH=BxG.
&c, &c,
we shall have, by addition,

A(D±F+H)=B(C+E+G);
and by separating the factors,

A : B : : C+£+G : D+F+H.

That is : In any number of 'proportions having the same
ratio, any antecedent will be to its consequent, as the sum of
the antecedents to the sum of the consequents.



240 FIRST LESSONS IN ALGEBRA.

Let us take, for example,

2 : 4 : : 6 : 12 and 1 : 2 : : 3 : 6, &c
Then, 2:4:: 6 + 3 : 12 + 6;

that is, 2 : 4 : : 9 : 18,

in which the ratio is still 2.

164. If we have four proportional quantities

A : B : : C : D, we have -^-=-77 ;

A G

and raising both members to any power, as n, we have

B n _D n

T n ~~c n '

and consequently

A n : B n : : C n : D\

That is : If four quantities are proportional, any like powers
or roots ivill be proportional.
If we have, for example,

2 : 4 : : 3 : 6,
we shall have 2 2 : 4 2 : : 3 2 : 6 2 ;
that is, 4 : 16 : : 9 : 36,

in which the terms are proportional, the ratio being 4.

165. Let there be two sets of proportions,

7? T)

A : B : : C : D, which gives -— =— ,

A O

F II
E : F : : G : H, „ „ ~E~7T'

Quest. — 163. In any number of proportions having the same ratio,
how will any one antecedent be to its consequent 1 — 164. In four pro-
portional quantities, how are like powers or roots 1



GEOMETRICAL PROGRESSION. 241

Multiply them together, member by member, we have
J^=^L w hich gives AE : BF : : CG : DH.

A Jit ULr

That is : In two sets of proportional quantities, the products
of the corresponding terms will be proportional.

Thus, if we have the two proportions





8


16 :


: 10 :


20


arid


3 :


4 :


: 6 :


8


we shall have


24 j


64 :


: 60 :


160



Geometrical Progression.

166. We have thus far only required that the ratio of
the first term to the second should be the same as that of
the third to the fourth.

If we impose the farther condition, that the ratio of the
second term to the third shall also be the same as that of the
first to the second, or of the third to the fourth, we shall have
a series of numbers, each one of which, divided by the
preceding one, will give the same ratio. Hence, if any
term be multiplied by this quotient, the product will be the
succeeding term. A series of numbers so formed is called
a geometrical progression. Hence,

A Geometrical Progression, or progression by quotients, is
a series of terms, each of which is equal to the product of



Quest. — 165. In two sets of proportions, how are the products of the
corresponding terms 1

21



242 FIRST LESSONS IN ALGRBRA.

that which precedes it by a constant number, which number
is called the ratio of the progression. Thus,

1 : 3 : 9 : 27 : 81 : 243, &c,

is a geometrical progression, which is written by merely
placing two dots between each two of the terms. Also,

64 : 32 : 16 : 8 : 4 : 2 : 1

is a geometrical progression, in which the ratio is one-half.

In the first progression each term is contained three times
in the one that follows, and hence the ratio is 3. In the
second, each term is contained one-half times in the one
which follows, and hence the ratio is one-half.

The first is called an increasing progression, and the
second a decreasing progression.

Let «, b, e, d, e, f, . . . be numbers in a progression by
quotients ; they are written thus :

a : b : c : d : e : f : g . . .

and it is enunciated in the same manner as a progression by
differences. It is necessary, however, to make the distinc-
tion, that one is a series of equal differences, and the other
a series of equal quotients or ratios. It should be remarked
that each term is at the same time an antecedent and a con-
sequent, except the first, which is only an antecedent, and
the last, which is only a consequent.



Quest. — 166. What is a geometrical progression 1 AVhat is the ratio
of the progression 1 If any term of a progression be multiplied by the


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Online LibraryCharles DaviesFirst lessons in algebra, embracing the elements of the science → online text (page 11 of 12)