Sidebilder
PDF
ePub

7. There are four numbers in arithmetical progression, of which the product of the two extremes is 45, and that of the means 77; what are the numbers?

Let x less extreme.

and y=common difference;

Then x, x+y, x+2y, x+3y, will be the four numbers, xx(x+3y)=x2+3xy=45

and

32

2(x+y)x(x+2y)=x2+3xy+2y2=77 question; Whence 2y2=77-45-32, by subtraction, and y2=· 2

16, by division.

Or y=16=4, by evolution;

Therefore x2+3xy=x2+12x=45, by the 1st equation; Also x2 + 12x+36=45+36=81, by completing the square,

And x+6=81=9, by the extraction of roots; ·Consequently x=9-6=3, and the numbers are 3, 7, 11, and 15.

8. To find three numbers in geometrical progression, whose sum shall be 14, and the sum of their squares

84.

Let x, y, and z be the numbers sought;
Then xz=y2, by the nature of proportion,

And

y z

{x2+12+2=14}, by the question;

But x+2=14-y, by the 2d equation.

And x2+2xz+z2-196-28y+y2, by squaring both sides, Or x2+z2+2y2-196-28y+y2, by putting 2y2 for its equal 2xz;

That is, x2+z2+y2=196-28y, by subtraction,

Or 196-28y84, by equality;

Hence y=

196-84

28

=4, by transposition and division.

16

Again, xz=y2=16, or x=— by the 1st equation,

16

[ocr errors]

And x+y+z == +4+z=14 by the 2d equation,

z

[ocr errors]

Or 16+4z + z2 14z, or z2-10z
Whence z2-10z+25=25-16

square,

=

-16,

9, by completing the

And z-5= √93, or z = 3 + 5 = 8, Consequently, x = 14-y —z=14 — 4 the numbers are 2, 4, 8.

-8=2, and

9. The sum (s) and the product (1) of any two numbers being given; to find the sum of the squares, cubes, biquadrates, &c. of those numbers.

Let the two numbers be denoted by x and y.

But

[blocks in formation]

xy=} by the question.

(x+y)2=x2+2xy+y2=s2, by involution, and x2+2xy+y2-2xy=82-2p, by subtraction. That is, x2+y2=s2—2p=sum of the squares. Again, (x2+y2)×(x+y)=(s2—2p)×8, by multiplication, Or x3+xyx(x+y)+y3=s3—28/2,

[ocr errors]

Or x3+sp+y3=s3 — 2sp, by substituting p for its equal xyx(x+y);

And therefore x3+y3—§3 38psum of the cubes. In like manner, (x3+y3)×(x+y)=(83—38f)×8, by mul tiplication,

Or x1+xyx(x2+y2)+y4=84—382/2,
s

[ocr errors]

Or x4+pX(+22)+y484382p, by substituting p (62-21) for its equal xyx(x2+y2);

And, consequently, x4+y454382 px (822/1)=81——— 482p+2p2= sum of the biquadrates, or fourth powers; And the sum of the nth. powers is sn nsn-2p + x.

[blocks in formation]
[ocr errors]
[ocr errors][merged small][merged small]

N-- 7 N- 8
.8 p4, Sc.

3

[ocr errors][merged small][ocr errors][ocr errors][merged small][merged small]

10. The sum (a) and the sum of the squares (b) of four numbers in geometrical progression being given; to find those numbers.

Let x and y denote the two means.

Then will

x2 42

Y

and be the two extremes, by the nature

of proportion.

[ocr errors]

Also, let the sum of the two means = s, and their product = = p.

Then will the sum of the two extremes — a—s by the question, and their product = p, by the nature of proportion.

[blocks in formation]

82-2/

(a—s)2 — 2p

304 74
+
y2 X2

x2 y2

}

by the last problem.

=82 + (a — 8)2 —4fb, by the

Again, + ——a—s, by the question.

3

y

[ocr errors]

Or x3+13=xy×(a—s)=p×(a—8).

But x3 y3:
3 = 53—3sp, by the last problem;
And therefore f×(a—»)=&3—38p, by equality,
Or pa-ps+3ps=pa+2ps=83,

[blocks in formation]

Whence 82+(as) 24/s2 + (α — 8)2 —

[blocks in formation]
[merged small][merged small][merged small][merged small][ocr errors][ocr errors]

by reduction.

a

2

[blocks in formation]

And, from this value of, all the rest of the quantities P,

by comp. the square,

2a

[ocr errors][merged small]

QUESTIONS FOR PRACTICE.

1. What two numbers are those whose sum is 20, and their product 36? Ans. 2 and 18. 2. To divide the number 60 into two such parts, that their product may be to the sum of their squares in the ratio of 2 to 5. Ans. 20 and 40.

3. The difference of two numbers is 3, and the difference of their cubes is 117; what are those numbers? Ans. 2 and 5.

4. A company at a tavern had 81. 153. to pay for their reckoning; but, before the bill was settled, two of them left the room, and then those who remained had 10s. apiece more to pay than before; how many were there in company?

Ans. 7.

5. A grazier bought as many sheep as cost him 601. and, after reserving 15 out of the number, he sold the remainder for 541. and gained 2s. a head by them; how many sheep did he buy? Ans. 75.

6. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number; what are those numbers? Ans. 3 and 18.

7. A person bought cloth for 331. 15s. which he sold again at 21. 8s. per piece, and gained by the bargain as much as one piece cost him; required the number of pieces.

Ans. 15.

8. What number is that, which, when divided by the product of its two digits, the quotient is 3; and, if 18 be added to it, the digits will be inverted?

Ans. 24.

9. What two numbers are those whose sum, multiplied by the greater, is equal to 77; and whose difference, multiplied by the lesser, is equal to 12?

Ans. 4 and 7, or 2 and 11,

[graphic]
[ocr errors]
[graphic]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
« ForrigeFortsett »