A series of numbers, such as 1, 2, 4, 8, 16, 32, 64, 128, 256, etc., in which any one of the terms is obtained by multiplying the preceding one by a constant quantity, is termed a Geometrical Progression. The constant quantity is termed the common Ratio, and is equal to any term divided by the preceding one. Thus in the above example 16/8 or 8/4 or 4/2 = 2, equals the common ratio of the above series. In the series, 1, 3, 9, 27, etc., we have for the ratio -

27/9 = 9/3 = 3/1 = 3;

which is the common ratio of this series. A geometrical series may be put thus:

Terms:

1

2,

3,

4;

Progress.:

1,

1 x 3,

1x3x3,

1x3x3x3;

or thus -

Terms:

1,

2,

3,

4:

Progress.:

1,

1 x 3,

1 x 32,

1x 33;

in which the common ratio, in this case 3, appears in each term and with an exponent which is equal to the number of terms preceding that in which it occupies a place.

If the first term be represented by a and the common ratio by r, then the following will represent any geometrical progression -

a, ar, ar2, ar3, ar4, etc. (124.)

For example, let a = 2 and r = 4; then the progression will be -

2, 8, 32, 128, 512, etc.

If r = unity, then when a = 2 the progression becomes -

2,2,2,2,2, etc.

If r be less than unity, then the progression will be a decreasing one.

For example, let a = 2 and r = 1/2. Then we have for the progression -

2, 1, 1/2, 1/4, 1/8, 1/16, etc

If the number of terms be represented by n, and the last by l, then the last term will be -

l= arn-1 (125.)

For example, let n equal 6, then the progression will be -

Terms:

1,

2,

3,

4,

5,

6;

Progress.:

a,

ar,

ar2,

ar3

ar4,

ar5;

in which the .exponent of the last term equals n - 1 = 6-1 = 5.

If 5 be put for the sum of a geometrical progression, we will have -

S = a + ar + ar2 +.....+ a rn-2 + a rn-1

Multiply each member by r, then -

Sr = ar + ar2 +.....+ arn-2 + arn-1 + arn.

Subtract the upper line from the lower; then -

Sr = ar + ar2 +.....+ arn-2 + a rn-1 + a rn,

S = a + ar + ar2 +.....+ arn-2 + arn-1_______

Sr - s = - a * * * +arn,

Sr - s= - a + arn,

S (r - 1) = - a + a rn = arn - a,

S = arn-a/r-1.

The last term (equation (125.)) equals l= arn-1, and since arn = r x arn-1 = rl, therefore -

S = rl-a (126.)

r - 1

Thus, to find the sum of a geometrical progression: Multiply the last term by the ratio; from the product deduct the first term, and divide the remainder by the ratio less unity.

For example, the sum of the geometrical progression -

5 = 1 + 3 + 9 + 27 + 81 + 243 + 729 = 1093

by actual addition.

To obtain it by the above rule -

430 Geometrical Progrression 338

the correct result.

If there be a decreasing geometrical progression, as 1, 1/3, 1/9,1/27, etc., in which the ratio equals -1/3, the sum will be -

S= 1+1/3+1 + 1/27+1/81 etc., to infinity.

Multiply this by 3, and subtract the first from the last -

3S=3+1/3+1/9+1/27+1/81 + to infinity.

S=1 +1/3 + 1/9 + 1/27 + 1/81 + to infinity.

2S = 3, or S = 1 1/2. In a decreasing progression let r, the common ratio, be represented by - (b less than c), and the first term by a, then the sum will be -

S = a+a b/c + a b2/c2 + a b3/c3 + , etc., to infinity.

Multiply this by b/c, and subtract the product from the above -

S b/c=a b/c + a b2/c2+a b3/c3 + etc., to infinity.

S = a + a b/c + b2/c2+a b3/c3 +to infinity.

S b/c = a b/c + a b2/c2+ a b3/c3 + to infinity.

S - S b/c = a* * *

Or- S(1-b/c)=a,

S = a/1-b/c (127.)

For example, let the first term of a geometrical progression equal 2, and the ratio equal 1/2, then the sum will be -

S = a/1-1/2 = 2/1/2 = 4/1 = 4

From this, therefore, we have this rule for the sum of an infinite geometrical progression, namely: Divide the first term by unity less the ratio.