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6. A and B 165 miles distant from each other set out with a design to meet; A travels 1 mile the first day, 2 the secon 1, 3 the third, and so on. B travels 20 miles the first day; 18 the second, 16 the third, and so on. How soon will they meet?

Ans. They will be together on the 10th day, and continuing that rate of travelling, they may be together again on the 3311 day. Let the learner explain how this can take place.

7. A person makes a mixture of 51 gallons, consisting of brandy, rum, and water

; the quantities of which are in arithmetical progression. The number of gallons of brandy and rum together, is to the number of gallons of rum and water together as 8 to 9. Required the quantities of each.

Let x = the number of gallons of rum and

y = the common difference. Then x — y, x, and x + y will express the three quantities.

8. A number consisting of three digits which are in arithmetical progression, being divided by the sum of its digits, gives a quotient 49; and if 198 be subtracted from the number, the digits will be inverted. Required the number.

9. A person employed 3 workmen, whose daily wages were in arithmetical progression. The number of days they worked was equal to the number of shillings that the second received per day. The whole amount of their wages was 7 guineas, and the best workman received 28 shillings more than the worst. What were their daily wages ?

Progression by difference is only a particular case of the series by difference, explained Arts. XL. and XLI. All the principles and rules of it may be derived from the formulas obtained there. It would be a good exercise for the learner to deduce these rules from those formulas.

XLVII. Progression by Quotient, or Geometrical Progression.

Progression by quotient is a series of numbers such, that if any term be divided by the one which precedes it, the quotient is the same in whatever part the two terms be taken. If the

series is increasing, the quotient will be greater than unity, if decreasing, the quotient will be less than unity.

The following series are examples of this kind of progression.

3, 6, 12, 24, 48....&c. 72, 24, 8,

7873 In the first the quotient (or ratio, as it is generally called,) is 2, in the second it is $. Let a, b, c, d, .... kl, be a series of this kind, and let

9 present the quotient.

Then we have by the definition,

re

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From these equations we derive

b =aq, c=b 9, d=cq, er dq... Putting successively the value of b into that of c, and that of c into that of d, &c., they become

b=aq, craq, d=aq", craq, ....l=aqa-, designating by n, the rank of the term l, or the number of terms in the proposed progression.

Any term whatever in the series may be found without finding the intermediate terms, by the formula

l-aq?

Example.

What is the 7th term of the series 3, 6, 12, &c. ?
Here a =

3,q=2, and 1=0.

1= 3 x 26 = 192.

n

Ans. 192

We may also find the sum of any number of terms of the progression

a, b, c, d, &c. If we add the equations

b=aq, c=bq, d=cq, e=dq .....l=kq, we obtain b+c+dtet...l= (a + b +c+dtet....k) q.

Observe that the first member is the sum of all the terms of the progression except the first, a, and the part of the second member enclosed in the parenthesis, is the sum of all the terms except the last, l; and this, multiplied by q, is equal to the first member.

Now putting S for the sum of all the terms, we have

b+c+dtet.....l=s-a a+b+c+dtet k=S-1. Hence we conclude that

S-a=(S-1) 9, which gives

ql-a
q 9-

S=

=

Example.

What is the sum of seven terms of the series

5, 15, 45, &c.

I=5 X 38 = 3645
S=
3 X 3645 – 5

= 5465.

3-1 The two equations l=aq*-, and S

91

a

contain all the relations of the five quantities a, l, q, n, and S; any three of which being given, the other two may be found. It would however be difficult to find n, without the aid of logarithms, which will be explained hereafter. Indeed logarithms will greatly facilitate the calculations in most cases of geometrical progression. Therefore we shall give but few examples, until we have explained them.

If we substitute a qn- in place of l, in the expression of S, it becomes

S=

a (q" — 1).

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When q is greater than unity, the quantity q” will become greater as n is made greater, and S may be made to exceed any quantity we please, by giving n a suitable value ; that is, by taking a sufficient number of terms.

is a fraction less than unity, the greater the quantity n, the smaller will be

1 the quantity q"Suppose q = -, m being a number greater

But if a

m

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Substituting

in place of q” in the expression of S, and it

m

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Changing the signs of the numerator and denominator, and multiplying both by m,

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It is evident that the larger n is or the more terms we take

a

in the progression, the smaller will be the quantity

and

ma

am

consequently the nearer the value of S will approach

т

from which it differs only by the quantity

a

a

(m— 1) mm But it can never, strictly speaking, be equal to it, for the quantity

will always have some value, however (m — 1) m"large n may be ; yet no quantity can be assumed, but this expression may be rendered smaller than it.

a m

m

The quantity

is therefore the limit which the sum

1 of a decreasing progression can never surpass, but to which the value continually approximates, as we take more terms in the series. In the progression

1, 3, 4, , ib, &c.

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- 2

1 x 2
1

1 Hence S=

2
-1
(2 — 1) X 2n-1

1 x 29 In this example the more terms we take, the nearer the sum of the series will approach to 2, but it can never be strictly equal to it. Now if we consider the number of terms infinite,

1 the quantity will be so small that it may be omitted

1 x 21-1 without any sensible error, and the sum of the series may be said to be equal to 2.

By taking more and more terms we approach 2 thus,

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