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Any number of quantities are said to be in Harmonical Progression when every three consecutive terms are in Harmonical Progression.

62. The reciprocals of quantities in Harmonical Progression are in Arithmetical Progression.

By definition, if a, b, c are in Harmonical Progression,

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63. Harmonical properties are chiefly interesting because of their importance in Geometry and in the Theory of Sound: in Algebra the proposition just proved is the only one of any importance. There is no general formula for the sum of any number of quantities in Harmonical Progression. Questions in H. P. are generally solved by inverting the terms, and making use of the properties of the corresponding A. P.

64. To find the harmonic mean between two given quantities.

Let a, b be the two quantities, H their harmonic mean; 1 1 1

then

a' H' b

are in A. P.;

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Here 6 is the 42nd term of an A. P. whose first term is

common difference; then

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65. If A, G, H be the arithmetic, geometric, and harmonic means between a and b, we have proved

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that is, G is the geometric mean between A and H.

From these results we see that

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(1).

(2).

.(3).

which is positive if a and b are positive; therefore the arithmetic mean of any two positive quantities is greater than their geometric

mean.

Also from the equation GAH, we see that G is intermediate in value between A and H; and it has been proved that A> G, therefore G>H; that is, the arithmetic, geometric, and harmonic means between any two positive quantities are in descending order of magnitude

66. Miscellaneous questions in the Progressions afford scope for skill and ingenuity, the solution being often neatly effected by some special artifice. The student will find the following

hints useful.

1. If the same quantity be added to, or subtracted from, all the terms of an A P., the resulting terms will form an A.P. with the same common difference as before. [Art. 38.]

2. If all the terms of an A.P. be multiplied or divided by the same quantity, the resulting terms will form an A.P., but with a new common difference. [Art. 38.]

3. If all the terms of a G.P. be multiplied or divided by the same quantity, the resulting terms will form a G.P. with the same common ratio as before. [Art. 51.]

4. If a, b, c, d... are in G.P., they are also in continued proportion, since, by definition,

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Conversely, a series of quantities in continued proportion may be represented by x, xr, xr2,

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Example 1. If a2, b2, c2 are in A. P., shew that b+c, c+a, a+b are in H. P.

By adding ab+ac + bc to each term, we see that

that is

a2+ab+ac+bc, b2+ba+bc+ac, c2+ca+cb + ab are in A.P.;

(a+b) (a+c), (b+c) (b+a), (c+a) (c+b) are in A, P..

..., dividing each term by (a+b) (b+c) (c+a),

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Example 2. If the last term, d the common difference, and s the sum of n terms of an A. P. be connected by the equation 8ds=(d+21)2, prove that

d=2a.

Since the given relation is true for any number of terms, put n=1; then a=l=s. 8ad=(d+2a)2,

Hence by substitution,

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Example 3. If the pth, qth, 7th, 8th terms of an A. P. are in G. P., shew that P-q, q-r, 7-8 are in G.P.

With the usual notation we have

=

a + (p-1) d _ a + (q − 1) d a+ (r-1) d
a + (q-1) d a + (r − 1 ) d ̄ ̄ a + (s − 1) d

... each of these ratios

=

=

=

=

[Art. 66. (4)];

{a+(p-1) d}-{a+(q-1) d} {a+(q-1) d} - {a+(r-1) d} {a+(-1) d} - {a+ (r−1) d} {a+(r-1) d} - {a+(s−1) d} p-q q-r

==

q-r 7-8

Hence p-q, qr, r-s are in G.P.

67. The numbers 1, 2, 3,...... are often referred to as the natural numbers; the nth term of the series is n, and the sum of

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68. To find the sum of the squares of the first n natural numbers.

Let the sum be denoted by S; then

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(n − 1)3 — (n − 2)3 – 3 (n − 1)3 — 3(n − 1) + 1 ; similarly (n-2)3 - (n − 3)3 = 3(n − 2)2 - 3(n − 2) + 1;

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69. To find the sum of the cubes of the first n natural numbers.

Let the sum be denoted by S; then

We have

S=13 + 23 +33 +

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1

(n − 1)* − (n − 2)1 = 4 (n − 1)3 — 6 (n − 1)2 + 4 (n − 1)−1; (n-2)* (n-3)=4(n-2)3-6 (n − 2) + 4 (n-2)-1;

Hence, by addition,

2

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324.3-6.32+4.3-1;

21' 4.23-6.2+4.2-1;
104.13-6.1+4.1-1.

n = 4S-6 (1 + 22 + ... + n3) + 4 (1 + 2 + .. + n) −n; ... 4S = n* + n + 6 (1a + 2a + ... + n3) − 4 (1 + 2 + ... + n)

12

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Thus the sum of the cubes of the first n natural numbers is equal to the square of the sum of these numbers.

The formulæ of this and the two preceding articles may be applied to find the sum of the squares, and the sum of the cubes of the terms of the series

a, a +d, a + 2d,........

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