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353. DEFINITION. When three quantities are in Geometrical Progression the middle one is called the geometric mean between the other two.

To find the geometric mean between two given quantities.

Let a and 6 be the two quantities; G the geometric mean. Then since a, G, b are in G.P.,

b G

a

each being equal to the common ratio;

.. G2=ab; whence

G=Vab. 354. To insert a given number of geometric means between two given quantities.

Let and b be the given quantities, n the number of means.

In all there will be n+2 terms; so that we have to find a series of n+2 terms in G. P., of which a is the first and b the last.

Let r be the common ratio; then

b=the (n+2)th term
=arn +1;
6

;

.. gone +1

a

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.(1). Hence the required means are ar, ar,...ar", where r has the value found in (1).

Example. Insert 4 geometric means between 160 and 5.

We have to find 6 terms in G. P. of which 160 is the first, and 5 the sixth. Let r be the common ratio;

then 5=the sixth term

=160r5;

1 32

1 whence, by trial, and the means are 80, 40, 20, 10.

.. 752

355. To find the sum of a number of terms in Geometrical Progression.

Let a be the first term, r the common ratio, n the number of terms, and s the sum required. Then

s=a+ar+are+ +arn-2+arn-1; multiplying every term by r, we have

rs=ar+ara + ...... tarn-2 + ar-1+arn. Hence by subtraction,

rs -- s=arn - O;
.. (r-1)= a(-1);
a(por – 1)

.(1).

p-1 Changing the signs in numerator and denominator [Art. 133.]

a(1 – gon)

....(2).

1-1 NOTE. It will be found convenient to remember both forms given above for s, using (2) in all cases except when r is positive and greater than 1.

Since arn-1=l, the formula (1) may be written

.. SE

S =

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2 Example 2. Sum the geries 1,

3' 2'

to 7 terms.
3
The common ratio= ; hence by formula (2)

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1. Find the 5th and gth terms of the series 3, 6, 12, ... 2. Find the 19th and 16th terms of the series 256, 128, 64,... 3. Find the 7th and 11th terms of the series 64, -32, 16,... 4. Find the gth and 12th terms of the series 81, -27, 9,...

1 1 1 5. Find the 14th and 7th terms of the series

64' 32' 16'"* 6. Find the 4th and 8th terms of the series •008, .04, -2, ...

Find the last term in the following series : 7. 2, 4, 8, ... to 9 terms. 8. 2, – 6, 18, ... to 8 terms. 9. 2, 3, 4),... to 6 terms.

10. 3, – 32, 33, ...

to 2n terms

1 11. X, X, X, ... to P terms. 12. X, 1,

,... to 30 terms. 13. Insert 3 geometric means between 486 and 6.

1 14. Insert 4 geometric means between and 128.

8

7 15. Insert 6 geometric means between 56 and

32 16. Insert 5 geometric means between

and 41. 81

16'

1

2

Find the last term and the sum of the following series : 17. 3, 6, 12,... to 8 terms. 18. 6, -- 18, 54, ... to 6 terms. 19. 64, 32, 16, ... to 10 terms. 20. 8:1, 2:7, 9, ... to 7 terms.

1 1 1 21.

to 8 terms. 22. 4), 1}, to 9 terms. 72' 24' 8°••• Find the sum of the series 1

1 1, to 6 terms. 24.

to 7 terms. 3'

2' 3'9' 2 1 5

1 1 25. to 6 terms. 26. 1,

to 12 terms. 5'2' g'

2'

2 1 1 27. 9, -6, 4, ... to 7 terms.

to 8 terms.

3' 6' 24' 29. 1, 3, 3, ... to

р

terms. 30. 2, – 4, 8, ... to 2p terms. 1

3 31. 1, to 8 terms. 32. Va, Nai, Ju',...to a terms.

28.

13.

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From this result it appears that however many terms be taken the sum of the above series is always less than 2. Also we see that, by making n sufficiently large, we can make the

1 fraction as small as we please. Thus by taking a sufficient number of terms the sum can be made to differ by as little as we please from 2.

In the next article a more general case is discussed.

2n-1

a(1 – por) 357. From Art. 355 we have s=

1

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a

=1-1-1 Suppose r is a proper fraction; then the greater the value of

arn n the smaller is the value of yon, and consequently of and

1 therefore by making n sufficiently large, we can make the sum of n terms of the series differ from by as small a quantity

1-r as we please.

This result is usually stated thus: the sum of an infinite number of terms of a decreasing Geometrical Progression is

1 or more briefly, the sum to infinity is ..

1

a

r

358. Recurring decimals furnish a good illustration of infinite
Geometrical Progressions.
Example. Find the value of .423.

-423= .4232323 ......
4 23 23
+

+
10 1000 100000

+

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which agrees with the value found by the usual arithmetical rule.

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