Find the common difference of the terms (212), and from that and either of the given terms find the required terms. Ans. 31, 47, and 632.

6. If the first term of an Arithmetical Progression be 36, the last term 150, and the number of terms 4; what are the intermediate terms? Ans. 74, and 112.

7. What is the sum of 1000 terms of an increasing Arithmetical Progression in which the first term is, and the common difference of the terms also? (211 & 215).

Ans. 250250.

GEOMETRICAL PROGRESSION.

(216.) A geometrical progression is a series of quantities in which each succeeding term has the same ratio to the term which immediately precedes it.

Thus 1, 2, 4, 8, 16, &c., is a Geometrical Progression in which each succeeding term is double the one which immediately precedes it; and the ratio of the progression is therefore 2.

And 27, 9, 3, 1, 4, &c., is a Geometrical Progression in which each succeeding term is one third of the one which immediately precedes it; that is, the ratio of the progression is.

State the Progression whose first term is 2, and ratio 3. State the Progression whose first term is 1, and ratio. State the Progression whose first term is, and ratio 4.

The first and last terms of a Progression are called the two extremes, and all the intermediate terms the means.

The principles of Geometrical Progression are contained in the following propositions.

(217.) The last term of a Geometrical Progression is equal to the first term × that power of the ratio which is expressed by the number of terms less one.

Thus if the first term be 1, and the ratio of the Progression 3, the series will be

1, 1x3, 1x3×3, 1×3×3×3, 1×3×3×3×3, &c. 1 X 33, 1 x 34, &c. ;

or 1, 1x3,

1 x 32,

in which it is plain that the last term, as 1 × 34, will always be equal to the first term, 1, multiplied by that power of the ratio, 3, which is expressed by the number of terms less one. From this it follows that

(218.) The last term of a Geometrical Progression the first term, gives that power of the ratio which is expressed by the number of terms less one.

Product of the two Extremes.

(219.) The product of the two extremes in a Geometrical Progression, is equal to the product of any two terms equidistant from them, or to the square of the middle term when the number of terms is odd.

Thus in the Progression

2, 2×3, 2×32, 2×33, 2×34,

the product 2×2×34 of the first and last terins is equal to that of the second and fourth terms which are equidistant from them, or to the square of the middle term 2×32.

From the preceding it also follows that

(220.) A geometrical mean between two given terms, is equal to the square root of the product of those terms.

For the product of the two given terms, regarded as the extremes of a Geometrical Progression, is equal to the square of the middle or mean term.

The Sum of all the Terms.

(221.) The sum of all the terms in a Geometrical Progression, is equal to the difference between the first term and the product of the last term × the ratio, the difference between the ratio and a unit.

For take the Progression

4, 4x3, 4x32, 4x33, 4x 34, the ratio being 3. Multiplying each term by the ratio, we obtain the series

4 x 3, 4x32, 4×33, 4x34, 4×35,

the sum of which is 3 times the sum of the first series. If the first series be subtracted from the second, the remainder will be

4×35-4, which must be twice the given series.

But this remainder is the difference between the first term, 4, and the product of the last term, 4 x 34, multiplied by the ratio 3; while 2 is the difference between this ratio and a unit. The sum of the series will therefore be found according to the proposition stated.

(222.) If the number of terms in a decreasing Geometrical Progression were infinite, that is, increased without limit, their sum would be equal to the first term the difference between the ratio and a unit.

Thus if the series 9, 3, 3, 3, and so on, in which the ratio is, were continued to an infinite number of terms; the

last term would be diminished without limit, that is, it would be 0; and by the preceding proposition, the sum of all the terms would be

` (9 —0 × }) ÷÷-− (1 −}) =9 ÷ 3=13.

On this principle we may also compute the value of a Repeating Decimal (89). For example, the repetend .4444, &c., is equal to

4

20+100+1000+10000, and so on without limit.

Now this is a decreasing Geometrical Progression in which the ratio is the sum of all the terms is therefore

;

which is the same value of the Repetend that would be found by the method before given (90).

The preceding propositions are to be applied to the following

EXERCISES.

the

1. The first term of a Geometrical Progression is 3, ratio of the progression is 5, and the number of terms 9; what is the last term? and the sum of all the terms?

The last term is equal to 3 × 58 (217); and the sum of all the terms is equal to (3× 58 × 5-3)÷(5—1) (221). Ans. 1171875; and 1464843. 2. The first term of a Geometrical Progression is 100, the ratio of the progression is, and the number of terms 7; what is the last term? and the sum of all the terms?

Ans. 100; and 149978. 729;

72

3. What is the ratio of a Geometrical Progression whose first term is 234, fourth term 1872, and number of terms 4?

In the preceding question we may readily find the third power of the ratio (218), and thence the ratio itself by evolution.

Ans. 2.

4. What is the second term in a Geometrical Progression whose first term is 5, and third term 1125 (220).

Ans. 75.

5. If the first and fourth terms of a Geometrical Progression be 34 and 918 respectively; what are the two intermediate terms?

Find the ratio of the progression (218), and from that and either of the given terms find the required terms.

Ans. 102 and 306.

6. If the first term of a Geometrical Progression be 15, the last term 960, and the number of terms 4; what are the intermediate terms? Ans. 60 and 240.

7. What is the sum of an infinite number of terms of a Geometrical Progression whose first term is 1000, and ratio? Ans. 2000.

8. How far would a person travel in 6 days, allowing him to go 40 miles the first day, and to diminish his rate in such a manner that each succeeding day's journey shall be of the one immediately preceding? Ans. 13167 miles.

9. If a body should move 2000 feet the first second, half that distance the next second, half the latter distance the next second, and so on, forever, what is the utmost distance it could go ? Ans. 4000.

10. If 11 yards of cloth were sold at 1 cent for the first yard. 3 for the second, 9 for the third, and so on, what would be the price of the last yard? and what would the whole amount to? Ans. $590.49; and $885.23.