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That is, If four quantities are proportional, any like power* >r root* will be proportional.

189. Let there be two sets of proportions,

B D

A : B :: C : D, which gives —=-771

A C

E : F :: G: H, * Multiply them together member by member, we have

J^-=~, which gives AE : BF :: CG : DH.

AUt O Cr

That is, In two sets of proportional quantities, the products of the '.orresponding terms will be proportional.

190. In the proportions which have been considered, it has only been 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 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 that which precedes it, by a constant number, which number is called the ratio of the progression. Thus, in the two series,

3, 6, 12, 24, 48, 96, . . .

64, 16, 4, 1, L, i, . . .

each term of the first contains that which precedes it twice, or is equal to double that which precedes it; and each term of the second is contained in that which precedes it four times, or is a fourth of that which precedes it. These are geometrical progressions. In the first, the ratio is 2; in the second, it is J. The first is called an increasing progression, the second a decreasing progression.

Let a, b, c, 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 distinction, 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 consequent, except the first, which is only an antecedent, and the last, which is only a consequent.

191. Let q denote the ratio of the progression

a : b : c : d . . .;

q being >1 when the progression is increasing, and 5,<1 when it is decreasing. We deduce from the definition the following equations:

b aq, c=bq = aq2, d=:cq~aq3, e—dq—aqi . . .;

and in general, any term n, that is, one which has n—1 terms before it, is expressed by aqn~l.

Let I be this term; we have the formula Z^oj"-1, by means of which we can obtain any term without being obliged to find all the terms which precede it. That is, the last term of a geometrical progression is equal to the first term multiplied by the ratio raised to a power whose exponent is one less than the number of terms.

1. Find the 5th term of the progression 2 : 4 : 8 : 16, &c, in which the first term is 2 and the common ratio 2.

5th term =2 x24=2 X 16 = 32.

2. Find the 8th term of the progression 2 : 6 : 18 : 54 . .«

8th term -2 x37=2x 2187=4374.

Ef. Find the 12th term of the progression

64 : 16 : 4 : 1 : — . . .

4

/1\U i> 1 1

12th term =64( —) =—=—= .

\4/ 411 48 65536

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

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

I denoting the nth term.

We have the equations (Art. 191),

b=aq, c = bq, d—cq, e—dq, . . . k~iq, l—kq;

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

b+c+d+e+ . . . +k+l=(a+b+c+d+ . . .

or, representing the required sum by S,

S— a=(S—l)q=Sq—Iq, or Sq—S=lq—a;

whence S=——

q-l

That is, to obtain the sum of the terms of a progression by quotients, multiply the last term by the ratio, subtract the first term from this product, and divide the remainder by the ratio diminished by unity.

1. Find the sum of eight terms of the progression

2 : 6 : 18 : 54 : 162 . . . : 2x37=4374,

g=jg-« 13122 -2 = 6560
q—l 2

2. Find the sum of the progression

2 : 4 : 8 : 16 : 32.

q-1 1

3. Find the sum of ten terms of the progression

2 : 6 : 18 : 54 : 162 . . . 2x39=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, $4 the third month, and so on, each succeeding payment being double the last; and what will be the last payment?

Ans. Debt, $4095; last payment, $2048.

5. A gentleman married his daughter on New Year's day, and gave her husband 1*. towards her portion, and was to double it on the first day of every month during the year: what was her portion? Ans. .£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 3rd, and so on to the last: what did he pay for the last bushel and for the ten bushels T

.Ans. Last bushel $196,83; total cost $295,24.

193. When the progression is decreasing, we have g,<l and Z<a; the above formula for the sum is then written under the form

——, in order that the two terms of the fraction may be 1-5

positive

By substituting <uf~l for I in the expression for S, it becomes

S=f£^L or S=^=X
q~1 l-q

1. Find the sum of the terms of the progression

32 : 16 : 8 : 4 : 2.

32-2x4" n

1—q 1 1.

T 2~

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

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S—g~fr _ 65536 4 _ 65536 655.35

~ 1-q ~ 3_ ~ 3 ~ 196608'

4

We perceive that the principal difficulty consists in obtaining the numerical value of the last term, a tedious operation, even when the number of terms is not very great.

alqn J)

194. Remark. If, in the formula S=————, we suppose

q=l, it becomes S=-^j-.

This result, which is sometimes a symbol of indetermination, is also often a consequence of the existence of a common factor (Art. 113), which becomes nothing by making a particular hypothesis respecting the given question. This, in fact, is the case in the present question; for the expression qn1 is divisible by q—1, (Art. 59), and gives the quotient

hence the value of S takes the form

S=aq"-1+aqn-2+af-3+ . . . +aq+a.

Now, making q—l, we have

S=a+<*+o+ . . . +a=na.

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