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then will a : c:: a : 12, or in the duplicate ratio of a : b

; a:d:: a3 : 63, or in the triplicate ratio of a:l; a: e:: a* : 6*, or in the quadruplicate ratio of 2 :b;

&c. &c. &c. &c. For, by Art. 98, a : 6:: 6 :c ::: d: e:: &c. &c. therefore, a :

b :: 6

:C; | and, by THEOREM 11, ao

6 : c. But, by THEOREM 3, 62 = aC; therefore, a?: 68 :: ac : c',

: 0, or a :c :: a? : 62.

:: C

:62

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(123.) The following Examples are designed to illustrate the application of the foregoing Theorems.

Er. 1. To divide the number 60 into two such parts, that the product shall be to the sum of the squares :: 2 : 5.

First, let x= one part; then 60-r= the other part,

(60-x) xx=60x—xo = the product; and x2 + (60-x2)=2x2+3600—120x=sum of the squares.

Hence, 600 — 12 : 2x? +3600—1200 :: 2 : 5, by the question ; consequently, by Theo. 1, (60.r-xo) x 5=(2x +3600—120r) x 2,

or 300x-5x'=4x2 + 7200-240x ; whence by transposition and division, x°-60.r= -800 ; therefore, x° -60+900=900-800=100, and x-30=+10;

or x=30+10=40 or 20 the

parts required. 12. The number 20 is divided into two parts, which are to each other in the duplicate ratio of 3:1. It is required to find a mean proportional between those parts.

First let x = greater part, then 20 x = lesser part;
Therefore by the question, X : 20 - 8:: 3% : 12 :: 9:1.
Hence, by Theorem 1, x = 180 -9r, or 10x = 180 ;
Therefore x = 18 greater part, and 20

20 - 18 = 2 Consequently by Theorem 3, a mean proportional between 18 and 2 is equal to 18 X 2 = 136 = 6 the number required.

3. If (a+x)2 : (-x)2 :: x + y: x — y, shew that a : 2 :: v2a-y : vy.

Here by expansici), ao + 2ax + x? ; a? 2ax +39 :: I+Y:X-4
But by Theorem 8, 2a + 2x® : 4ax :: 2x : 2y.
Divide then by 2, and a2 + x 2 : 2ax :: 3:4;

lesser part.

:

dx = cy.

Therefore, by Thecrem 1, (a® + x2) x y = 2ax Xx = 2a x x®.
Hence, by Theorem 2, a? + x2 : x2 :: 2a : y.

By Theorem 6, a® : x2 :: 20–4:Y;
and by Theorem 11, (n being 1) a :x:: 2a-y: wy.

4. If r:y in the triplicate ratio of a : b, and a :b :: $c + x:
Sa+y, shew that dr = cy.
Here since x : Y

ad'

:6, and by Theorem 11, as : 23 C+* :

d +y;
therefore by Theorem 5,4 :y:: ctx:d+y,

or ctx :d+y :: X: Y,
and by Theorem 4,c+x:x :: d+y: Y;
:. by Theorem 6, c: x :: d

y ;
and by Theorem 1,
5. There are two numbers, the product of which is 24, and the
difference of their cubes : cule of their difference :: 19:1. What
are the numbers ?
Let x =

greater number, and y= less number.
Then, by the question, xy 24, and 2:— 43 : (xy)3 :: 19:1.
Or by expansion, 23 – 73 : 23-3z'y +3.ry* - ys (r-y:: 19:1.
And by Theorem 8, 3.xoy - 3xy? : (ry)? :: 18:1,

or 3xy x (x - y): (-y)3 :: 18:1.
Divide now by r — y, then will 3xy : (xy)? :: 18:1;

24; ::.72 : (1-y)? :: 18: 1. Hence, by Theorem 1,18 (r-y) = 72,

or (r-y) = 4;

i.t-y = 2.

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but xy

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6. It is required to divide the number 24 into two such parts, that their product shall be to the sum of their squares :: 3 : 10.

Ans. 18 and 6. 7. There are two numbers which are to each other as 3 : 2. If 6 be added to the greater, and subtracted from the less, the sum and remainder will be to each other :: 3:1. Wliat are those two numbers ?

Ans. 24 and 16. 8. There are two numbers which are to each her the duplicate ratio of 4:3, and 24 is a mean proportional between them. What are those two numbers ?

Ans. 32 and 18.

9. If x : ye :: 36 : 25, and 2x + y : x + 2 in a ratio compounded of the ratios of 17 :: 2 and 2:7; what are the two numbers ?

Ans. 12 and 10. 11. If a +2:0 - :: 9:5, shew (by Theor. 8.) that a : 2 :: 7:2.

10. There are two numbers whose product is 135, and the difference of their squares is to the square of their difference :: 4:1. What are these numbers ?

Ans, 15 and 9.

PROGRESSION.

(124.) When three or more quantities are in continued pro. portion, such that, as the first is to the second, so is the second to the third, so is the third to the fourth, so is the fourth to the fifth, &c. the series is called a progression, and is either an Arithmetical or a Geometrical Progression, according as the several terms of the series are in Arithmetical or Geometrical Proportion.

And any progression is said to be ascending when the terms of the series increase, but to be descending when they decrease.

Arithmetical Progression. (125.) The chief properties of an arithmetical series are the following:

1. The sum of the extreme terms is equal to the sum of any two means equidistant from them, or to twice the middle term when the number of terms is odd.

2. The difference of the extremes, divided by 1 less than the number of terms, is equal to the common difference.

3. The difference of the extremes is equal to the common difference multiplying 1 less than the number of terms.

4. The sum of the series is equal to half the number of terms multiplying the sum of the extremes. (126.) If, therefore, in any Arithmetical Series

The first term be denoted by
The last term by ..
The number of terms by
The common difference by .. d
..nd the sum of the series by

a

n

S

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3. Given a=10, =80, and n=9, to find s.

in(a+x)={x9(10+80)=405. Ans. 4. Given a=1, n=100, and d=5, to find x in an ascending series.

a+d(n-1)=1+5 x 99=496. Ans. 5. Given as =1, n=100, and d=2, to find x in a descending series.

Ans. -197 6. Given a=245, n=13, and d=10, to find % in a descending serie

Ans, 125. 7. Given a=49, nsa 50, and d = , to find , in a descending series.

Ans. 683. Note.-If n, d, and z were given to find a in either series, it is manifest that, by reversing the series, a may be substituted for z, and z for a, since the first term of any ascending series is the last term of the same series descending. 8. Given n=13, d 10, and %=125, in a descending series, to

find a.

By reversing the series, it will be, a=125, n=13, and d=10, to find x in an ascending series.

But a+d(n-1)=125+10(13-1)-245. Ans. 9. Given n=7, d=3, and x=90, in an ascending series, to

Ans, 72. 10. It is required to find four numbers in arithmetical progression, such that their common difference may be 4, and their continued product 176985.

Ans. 15, 19, 23, and 27.

of

find a.

11. The sum of four numbers in arithmetical progression is 56, and the sum of their squares is 864 ; what are the numbers?

Ans, 8, 12, 16 and 20

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Promiscuous Questions. 1. What is the sum of a million terms of the series 1, 2, 3, &c. ?

Ans. 500,000,500,000. 2. What is the sum of 500 terms of the series 1,3,5, &c. ?*

Ans. 250,000. 3. How often does the hammer of a common clock, keeping correct time, strike the bell in 30 complete days?

Ans. 4680 times. 4. If the series 1, 4, 7, &c. be carried to 91 terms, how much will the last term exceed the third ?

Ans. 269. 5. What must be paid for 80 trees, the price of the first being fixed at 5s. and of the last at 181. all the intermediate prices being in arithmetical progression ?

Ans. 7301. 6. What is the number of acres in an estate, which, if sold on the principles of arithmetical progression, 6d. being given for the first acre, and gd, for the second, the last acre would be valued at 1001. ?

Ans. 7999 7. What sum will be just sufficient to purchase the estate is Question 6th, with all its conditions ?

Ans. 400,0491. 198. 9d. 8. If, on the 1st of January, 200 copies of a new work be ‘isposed of, and the sale decrease in arithmetical progression to the -nd of the month, in such manner that on the 31st only 20 copies are sold, what is the common difference of the terms >

Ans. 6. 9. Let 10 and 70 be the two extremes of an arithmetical series, and three the common difference. Required the number of terms, and the sum of the series.

Ans. (n) = 21; (s) = 840. 10. Wbat debt can be discharged in a year, by weekly payments in arithmetical progression ; the first payment is., the last 25.35.; and what is the difference between each payment ?

Sum...... £135 4 0.

2 C 11. Required the rule for solving the following question :-A person takes out of his pocket, at ten several times, so many different numbers of pounds, every one exceeding the former by 2; the last was 23. Required the sum taken at first.

12. Let the last term of an increasing arithmetical series be 19, the common difference 2, and the number of terms 9. Required the sum of the series.

Ans. 99. 12 Find the last term of an arithmetical series ; the sum being 99, the number of terms 9, and the first term 3.

ins. 19.

Ans. { Difference.

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