Charles Davies.

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

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Online LibraryCharles DaviesFirst lessons in algebra, embracing the elements of the science → online text (page 12 of 12)
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ratio, what will the product be 1 If any term be divided by the ratio,
what will the quotient be 1 How is a progression by quotients written 1
Which of the terms is only an antecedent 1 Which only a consequent ?
How may each of the others be considered 1

GEOMETRICAL PROGRESSION. 243

167. Let q denote the ratio of the progression

a : b : c : d . . . ;

q being >1 when the progression is increasing, and ^<1
when it is decreasing. Then, since

b c d e

T= q ' T= ? ' T =? ' T= ? ' &c '

we have

b = aq, c = bqz=aq 2 , d=cq=aq 3 , e=dq=zaq i ,
' f—eq = aq 5 . . .;

that is, the second term is equal to aq, the third to aq 2 , the
fourth to aq 3 , the fifth to aq^, &c ; and in general, any term
n, that is, one which has n— 1 terms before it, is expressed
by aq n ~ l .

Let I be this term ; we then have the formula

l—aq n - 1 ',

by means of which we can obtain any term without being
obliged to find all the terms which precede it. Hence, to
find the last term of a progression, we have the following

RULli

I. Raise the ratio to a power whose exponent is one less titan
the number of terms.

II. Multiply the power thus found by the first term : the
product will be the required term.

Quest. — 167. By what letter do we denote the ratio of the progres-
sion 1 In an increasing progression is q greater or less than 11 In a
decreasing progression is q greater or less than 11 If a is the first term
and q the ratio, what is the second term equal to 1 What the third 1
"What the fourth 1 W T hat is the last term equal to 1 Give the rule for
finding the last term.

244 FIRST LESSONS IN ALGEBRA.

EXAMPLES.

1. Find the 5th term of the progression

2 : 4 : 8 : 16 . .
in which the first term is 2 and the common ratio 2.
5th term=2 x2 4 =2 X 16 = 32 Ans.

2. Find the 8th term of the progression

2 : 6 : 18 : -54 . . .

8th term=2 X 3 7 =2 X 2187=4374 Ans.

3. Find the 6th term of the progression

2 : 8 : 32 : 128 . . .
6th term =2 x 4 5 =2 X 2048=4096 An s .

4. Find the 7th term of the progression

3 : 9 : 27 : 81 . . .

7th term=3 x 3 6 =3 X 729=2187 Ans.

5. Find the 6th term of the progression

4 : 12 : 36 : 108 . . ,
6th term =4 X 3 5 =4 X 243 = 972 Ans.

6. A person agreed to pay his servant 1 cent for the first
day, two for the second, and four for the third, doubling
every day for ten days : how much did he receive on the
tenth day? Ans. \$5,12

GEOMETRICAL PROGRESSION. 245

7. What is the 8th term of the progression

9 : 36 : 144 : 576 . . .
8th term =r 9 x4 7 = 9x16384 = 147456 Ans.

8. Find the 12th term of the progression

64 : 16 : 4 : 1 : 4" • • •
4

/ 1 \ u 4 3 1 1

12th term = 64( — ) = = — xs Ans.

\4/ 4 11 4 8 65536

168. We will now proceed to determine the sum of n
terms of the progression

a : b : c : d : e : f : . . . i i : k : I;

I denoting the nth term.

We have the equations (Art. 167),

b=aq, c—bq, d = cq, e=dq, . . . k=ziq, l=zkq;

and by adding them all together, member to member, we
deduce

Sum of 1st members. Sum of 2nd members.

b+c+d+e+ . . . +k + l=(a+b + c+d+ . . . +t + %;

in which we see that the first member wants the first term
a, and the polynomial within the parenthesis in the second
member wants the last term I. Hence, if we call the sum
of the terms S, we have

S — a=(S — l)q=Sq— Iq, or Sq—S=zlq—a;

whence S = — — — -.

2- 1

246 FIRST LESSONS IN ALGEBRA.

Therefore, to obtain the sum of the terms of a geometrical
progression, we have the following

RULE.

I. Multiply the last term by the ratio.

II. Subtract the first term from the product.

III. Divide the remainder by the ratio diminished by unity ',
and the quotient will be the sum of the series.

1 . Find the sum of eight terms of the progression

2 : 6 : 18 : 54 : 162 . . . 2x3 7 = 4374.

s= J^ = 13122-3 =656()
£ — 1 2

2. Find the sum of the progression

2 : 4 : 8 : 16 : 32.

q-\ 1

3. Find the sum of ten terms of the progression

2 : 6 : 18 : 54 : 162 . . . 2x3 9 = 39366.

Ans. 59048.

4. What debt may be discharged in a year, or twelve
months, by paying \$1 the first month, \$2 the second month,

Quest. — 168. Give the rule for finding the sum of the series. What
is the first step 1 What the second 1 What the third ?

GEOMETRICAL PROGRESSION* 247

\$4 the third month, and so on, each succeeding payment
being double the last ; and what will be the last payment ?

A ( Debt, . . \$4095.
" \ Last payment, \$2048.

5. A gentleman married his daughter on New Year's day,
and gave her husband Is. towards her portion, and was to
double it on the first day of every month during the year :
what was her portion? Arts. j£204 15s.

6. A man bought 10 bushels of wheat on the condition
that he should pay 1 cent for the 1st bushel, 3 for the second,
9 for the third, and so on to the last : what did he pay for
the last bushel and for the ten bushels ?

A i Last bushel, \$196,83.
nS ' \ Total cost, \$295,24.

7. A man plants 4 bushels of barley, which, at the first
harvest, produced 32 bushels ; these he also plants, which,
in like manner, produce 8 fold ; he again plants all his crop,
and again gets 8 fold, and so on for 16 years : what is his
last crop, and what the sum of the series 1

Ans < Last, 1407374883553285w.
nS ' ( Sum, 160842843834660.

169. When the progression is decreasing, we have
§'<1 and Z<« ; the above formula

q— 1

for the sum is then written under the form

\-q
in order that the two terms of the fraction may be positive.

Quest. — 163. What is the formula for the sum of the series of a
decreasing progression 1

248 FIRST LESSONS IN ALGEBRA.

1. Find the sum of the terms o£ the progression

32 : 16 : 8 : 4 : 2.

32-2x4" qi

l—q 1 1

~2 ~2~

2. Find the sum of the first twelve terms of the progression

1

64 : 16 : 4 : 1 : — :... : 64^— \ ,

65536

64 - -^— Xt 256 - ]

<z— Z^_ 65536 4_ 65536 _o 5 , 65535

1— q 3 3 196608

T

Remark. — 170. We perceive that the principal diffi-
culty consists in obtaining the numerical value of the last
term, a tedious operation, even when the number of terms
is not very great.

3. Find the sum of 6 terms of the progression

512 : 128 : 32 . . .

Arts. 682 j.

4. Find the sum of seven terms of the progression

2187 : 729 : 243 . . .

Ans. 3279.

5. Find the sum of six terms of the progression

972 : 324 : 108 . . .

Ans. 1456.

6. Find the sum of 8 terms of the progression

147456 : 36864 : 9216 . . .

Ans. 196605.

GEOMETRICAL PROGRESSION. 249

Of Progressions having an infinite number of terms.

171. Let there be the decreasing progression
a : b : c : d : e : f : . . .
containing an indefinite number of terms. In the formula

l-q
substitute for I its value aq n ~ l (Art. 167), and we have

~ l-q '

which represents the sum of n terms of the progression.
This may be put under the form

s= JL- -<-

l-q l-q

Now, since the progression is decreasing, q is a proper
fraction ; and q n is also a fraction, which diminishes as n
increases. Therefore, the greater the number of terms we
take, the more will x q n diminish, and consequent-
ly the more will the partial sum of these terms approximate

to an equality with the first part of S, that is, to .

Finally, when n is taken greater than any given number, or

H= infinity, then xq n will be less than any given

a

number, or will become equal to ; and the expression

will represent the true value of the sum of all the terms of
the series. Whence we may conclude, that the expression

250 FIRST LESSONS IN ALGEBRA.

for the sum of the terms of a decreasing progression, in which
the number of terms is infinite, is

S=- "

l-q

That is, equal to the first term divided by 1 minus the ratio.

This is, properly speaking, the limit to which the partial
sums approach, by taking a greater number of terms in the
progression. The difference between these sums and

can become as small as we please, and will only

\-q

become nothing when the number of terms taken is infinite.

EXAMPLES.

1 . Find the sum of

1111 . „ .

1 : T : y : 27 : 8i t0infinit ^

We have for the expression of the sum of the terms
a 13

"'-'">4

Ans.

The error committed by taking this expression for the
value of the sum of the n first terms, is expressed by

; it becomes

A 3 / 2 . ;

First take n=5 ; it becomes

_3_/J_\ 5 _ 1 1

2 \ 3 / 2 . 3 4 ~~ 162

Quest. — 165. When the progression is decreasing and the number of
terms infinite, what is the value of the sum of the series 1

GEOMETRICAL PROGRESSION. 251

"When n=6, we find

3 / 1 \ 6 _ 1 1 _ 1

T\T/ ~~ 162 X Y"~ 486 '

3

Whence we see that the error committed, when — is

taken for the sum of a certain number of terms, is less in
proportion as this number is greater.

2. Again take the progression

2 " 4 * 8 ' 16 * 32 ' C * ' ' *
We have S= = —-=2. Ans.

3. What is the sum of the progression

^To' Too - ' Tooo' ToW J &Cj toinfinity '
10

172. In the several questions of geometrical progres-
sion there are five numbers to be considered :

1st. The first term, a.

2nd. The ratio, q.

3rd. The number of terms, n.

4th. The last term, I.

5th. The sum of the terms, S.

Quest. — 166. How many numbers are considered in geometrical pro-
gression 1 What are they 1

252 FIRST LESSONS IN ALGEBRA.

173. We shall terminate this subject by the question,

To find a mean proportional between any two numbers,
as m and n.

Denote the required mean by x. We shall then have
(Art. 156),

sc 2 =z mXn,

and hence x = -\/mxn.

That is, Multiply the two numbers together, and extract the
square root of the product.

1. What is the geometrical mean between the numbers
2 and 8 1

Mean == y8x2 = V 16 = 4 Ans.

2. What is the mean between 4 and 16 ? Ans. 8.

3. What is the mean between 3 and 27 ? Ans. 9.

4. What is the mean between 2 and 72 ? Ans. 12.

5. What is the mean between 4 and 64 ? Ans. 16.

Quest. — 167. How do you find a mean proportional between two
numbers' 1

THE END.

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