This operation may be performed more simply, as follows:

Let S .253131. . . . Multiply by 100, in order to remove the decimal point to the commencement of the first period of decimals, we have

100S-25.3131...

Again, multiplying by 100 to remove the decimal point to the commencement of the second period of decimals, we have

10000S-2531.3131.

Subtracting the preceding equation from the last, we get 9900S=2506; .. S=3388.

ART. 302. Three or more quantities are said to be in Harmonical Progression, when their reciprocals are in arithmetical progression.

Thus,

1

1, 1, 1, 4, &c.; and 1,,,, &c.

are in harmonical progression, because their reciprocals 1, 3, 5, 7, &c.; and 4, 31, 3, 21, &c.

are in arithmetical progression.

ART. 303. PROPOSITION.-If three quantities are in harmonical progression, the first term is to the third, as the difference of the first and second, is to the difference of the second and third,

For if a, b, c, are in harmonical progression,

Dividing both sides by a-b, and by a, we have

Therefore, a Harmonical Progression is a series of quantities in harmonical proportion (Art. 280); or such that if any three consecutive terms be taken, the first is to the third, as the difference of the first and second is to the difference of the second and third.

From this proposition it follows, that all problems with respect to numbers in harmonical progression, may be solved by inverting them, and considering the reciprocals as quantities in arithmetical progression. This renders it unnecessary to give any special rules for the solution of problems in harmonical progression.

1. Given the first two terms of a harmonical progression, a and b, to find the nth term.

1

Let I be the nth term, then (Art. 302), and are the first

α

b

two terms of an arithmetical progression, and it is required to

By means of this formula, when any two successive terms of a harmonical progression are given, any other term may be found. 2. Insert m harmonic means between a and b.

Here, if d be the common difference of the reciprocals of the terms, we have

1 =2+(n−1)d, and d—_a_b

ს α

d=

a-b
(n-1)ab (m+1)ab

whence the arithmetical progression is found; and by inverting its terms, the harmonicals are also found.

3. Insert two harmonic means between 3 and 12.

4

Ans. 4 and 6.

4. Insert two harmonic means between 2 and 1.

Ans.and.

5. The first term of a harmonic series is 2, and the 6th; find the intermediate terms. 1 Ans.,, 10.

6. a, b, c, are in arithmetical progression, and b, c, d, are in harmonical progression; prove that a:b::c: d.

PROBLEMS IN ARITHMETICAL AND GEOMETRICAL PRO

GRESSION.

ART. 304. The sum of five numbers in arithmetical progression is 35, and the sum of their squares 335; find the numbers. Ans. 1, 4, 7, 10, 13.

Let x-2y, x-y, x, x+y, x+2y, be the numbers.

2. There are four numbers in arithmetic progression, and the sum of the squares of the extremes is 68, and of the means 52; find them.

Ans. 2, 4, 6, 8.

Let x-3y, x-y, x+y, x+3y, be the numbers.

SUGGESTION. When the number of terms in an arithmetic progression is odd, the common difference should be called y, and the middle term x; but when the number of terms is even, the common difference must be 2y, and the two middle terms x-y, and xy.

3. The sum of 3 numbers in arithmetical progression is 30, and the sum of their squares 308; find them. Ans. 8, 10, 12. 4. There are 4 numbers in arithmetical progression, their sum Ans. 2, 5, 8, 11.

Y

is 26, and their product 880; find them. 5. There are 3 numbers in geometrical progression, whose sum is 31; and the sum of the 1st and 2nd: sum of 1st and 3rd :: 3:13; find them. Ans. 1, 5, 25.

6. The sum of the squares of three numbers in arithmetic progression is 83; and the square of the mean is greater by 4 than the product of the extremes. Required the numbers.

7. Find 4 numbers in arithmetical progression, such that the product of the extremes =27; of the means =35.

8. There are 3 numbers in arithmetical progression, whose

2

257

ARITHMETIC AND GEOMETRIC PROGRESSION.

sum is 18; but if you multiply the first term by 2, the second by
3, and the third by 6, the products will be in geometrical pro-
gression; find them.

Ans. 3, 6,

9.
9. The sum of the 4th powers of three successive natural num-
bers is 962; find them.
Ans. 3, 4, 5.
10. The product of four successive natural numbers is 840

find them,

11.

The

y- 37

Ans. 4, 5, 6, 7.

ener Hey! ·2=411⁄2=572 412-572 product of four numbers in arithmetical progression

is 280, and the sum of their squares 166; find them.

12. The sum of 9 numbers in arithmetical progression is 45, and the sum of their squares 285; find them.

Ans. 1, 2, 3, &c., to 9.

13. The sum of 7 numbers in arithmetical progression is 35,

and the sum of their cubes 1295; find them.

と

Ans. 2, 3, &c., to 8.

14. Prove that when the arithmetical mean of two numbers is to the geometric mean:: 5:4; that one of them is 4 times the other. 2

15. The sum of 3 numbers in geometrical progression is 7;
and the sum of their reciprocals is; find them. Ans. 1, 2, 4.

16 There are 4 numbers in geometrical progression, the sum
of the first and third is 10, and the sum of the second and fourth
is 30; find them.
Ans. 1, 3, 9, 27.
17. There are 4 numbers in geometrical progression, the sum
of the extremes is 35, the sum of the means is 30; find them.
Ans. 8, 12, 18, 27.

18. There are 4 numbers in arithmetical progression, which
being increased by 2, 4, 8, and 15 respectively, the sums are in
geometrical progression; find them.
Ans. 6, 8, 10, 12.
19. There are 3 numbers in geometrical progression whose
continued product is 64, and the sum of their cubes 584; find
them.
Ans. 2, 4, 8.

SUGGESTION. In solving difficult problems in geometrical progression,
instead of denoting the terms by x, xy, xy2, &c., it is sometimes prefer-
able to express them by other forms. Thus, three terms may be ex-
pressed by x, √ xy, y, or, x2, xy, y2; four terms by

x2

these cases the product of the first and third of any three consecutive
terms, is equal to the square of the middle term.

X-24-22

3

CHAPTER IX.

PERMUTATIONS, COMBINATIONS, AND BINOMIAL THEOREM.

ART. 305. The different orders in which quantities can be arranged, are called their Permutations. Quantities may be arranged in sets of one and one, two and two, three and three, and so on. Thus, if we have three quantities, a, b, c, we may arrange them in sets of one, of two, or of three, thus:

REMARK.-Some writers, confine the term permutations to the class where the quantities are taken all together, and give the title of arrangements, or variations, to those groups of one and one, two and two, three and three, &c., in which the number of quantities in each group is less than the whole number of quantities.

ART. 306. To find the number of permutations that can be formed out of n letters, taken singly, taken two together, three and r together.

together.

Let a,

b, c, d,

k, be the n letters; and let P1 denote the whole number of permutations where the letters are taken singly; P2 the whole number of permutations taken 2 together. and P, the whole number of permutations taken

r together.

2

The number of permutations of n letters taken singly, or one and one, is evidently equal to the number of letters, that is n; therefore,

The number of permutations of n letters, taken two together, is n(n-1). For since there are n quantities

if we remove a, there will remain (n-1) quantities,

Writing a before each of these (n-1) quantities, we shall

That is, (”—1) permutations in which a stands first.