A Decreasing Progression with an Infinite Number of

Terms.

§ 354. When the ratio of a geometrical progression is a proper fraction, the succeeding terms will continually diminish; and if the number of terms be supposed infinite, the last terin will be 0.

1

In the series 3,, 1, 27, 31, and so on, in which the ratio of progression is, the terms continually diminish, and the series would terminate in 0 if the number of terms were infinite.

§ 355. The sum of an infinite number of terms in a decreasing geometrical progression, is equal to the first term divided by the difference between the ratio and a unit.

The sum of the series

11 1 1

3, 3;3; 27: 81, and so on, to an infinite number of terms, according to § 353, is

(3-the last term X the ratio)÷(1—}}.

But the last term would be 0 (§ 354); hence the sum of the infinite series is 3-(1-3)=3+3=1.

On the principle just demonstrated we may also compute the value of a repeating decimal.

For example, take the repetend .4,

This decimal is equal to

or .4444 &c.

10+100+1000+10, and so on, without end.

We have then .4 equal to the sum of an infinite geometrical series of which the first term is 4 and the ratio 0.

Hence 4(1-1)=4÷8==}.

49 10 10

This value, it will be observed, is the same that would be found by the method before given, (§ 328).

EXERCISES.

1. What is the last term in a geometrical progression of which the first term is 21, the ratio 3, and the number of terms 5?

2. What is the sixth term in a geometrical which the first term is 7 and the ratio ?

Ans. 202. progression of Ans. 29

3. What will be the first term of a geometrical progression, if the ratio be 2, the number of terms 9, and the last term 1280?

Ans. 5.

4. What is the second or mean term in a geometrical progression of which the first and third terms are 64 and 30₫ respectively? Ans. 13.

5. What is the last term, and the sum of all the terms, in a geometrical progression whose first term is 23, ratio 5, and number of terms 4? Ans. 2875, and 3588. 6. What is the sum of the first six terms in a geometrical progression whose first term is 100, and ratio?

Ans. 273 161 7. What is the sum of an infinite number of terms in a geometrical progression whose first term is 1000, and ratio ?

Ans. 2000.

8. What is the sum of the series 5, 1, 1, and so on to an infinite number of terms? Ans. 64.

9. What is the sum of the infinite series 100, 50, 25, &c.?

Ans. 200. 10. What is the difference between the arithmetical and the geomètrical mean between the numbers 100 and 1000? Ans. 233.773'.

11. If 9 acres of land be sold at $1 for the first acre, $2 for the 2d, $4 for the 3d, and so on, to the last, what would the last acre amount to? Ans. $256.

12. A has $400, B $900, and C a sum which bears the same ratio to A's that B's does to C's: how many dollars then has C? Ans. $600.

13. If 11 yards of cloth were sold at 1 cent for the first yard, 3 for the 2d, 9 for the 3d, and so on to the last, what would be the price of the last yard? and what would the whole amount to? Ans. $590.49; and $885.73.

14. How far would a person travel in 6 days, allowing him to go 40 miles the first day, and to diminish his rate of travelling in such a manner that each succeeding day's journey shall be of the one immediately preceding. Ans. 131 miles.

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128

15. Allowing a person to commence trading on a capital of $1000, and to increase it by of itself each year for 10 years, what would then be the amount of his capital?

Ans. $7450.580'.'

16. If 50 acres of land were to be sold at 1 cent for the first acre, 2 for the second, 4 for the third, and so on, what would the 50 acres amount to? Ans. $11258999068426.23.

PROGRESSIONS APPLIED TO INTEREST.

I. Simple Interest.

§356. A Principal at simple interest forms the first term of an Arithmetical Progression-in which the common difference is the Interest for one year-the number of terms one more than the number of years for which interest is computed-and the last term the Amount of principal and interest.

For example, the Interest of $100, at per cent, being $6 for 1 year, $12 for 2 years, $18 for three years, and so on,— We have the Principal, and the Amounts for one, two, three, &c. years, in the terms of the Arithmetical Progression,

$100, $106, $112, $118, &c.

The common difference of these terms is $6, the Interest for one year.

II. Compound Interest.

§ 357. A Principal at compound interest forms the first term of a Geometrical Progression-in which the ratio is the Amount of $1 for one year or period of interest-the number of terms one more than the number of years or periods of interest-and .the last term the Amount of principal and interest.

The amount of $100, at 6 per cent, for the 1st year, would be $106, and this would form the principal for the 2d year, (§ 282).

But 106 is equal to 100X1.06; and $1.06 is the Amount of $1 for one year; that is, the principal for the 1st year 1.06 produces the Amount for the 1st or the principal for the 2d year.

In like manner, $106X1.06 or $100X1.06X1.06 is the Amount for the 2d or the principal for the 3d year; $100×(1.06)3 is the Amount for the 3d or the principal for the 4th year, &c.

We have then the Principal, and the Amounts for one, two, three, &c. years, in the terms of the Geometrical Progression, $100, $100x1.06, $100×(1.06), $100x(1.06) 3, &c.

The ratio of this progression is 1.06, the Amount of $1 for one year, at the given rate per cent.

The given Principal subtracted from the Amount for the last year, will leave the Compound Interest.

Tabular Method of Calculating Compound Interest.

§ 358. In finding the Amount from the Principal, or the principal from the amount, in Compound Interest, on the principles of geometrical progression,-the ratio must be raised to that power which is expressed by the number of years, or periods.

The TABLE on the next page contains these powers to the 50th order, for various rates of interest; and thus supersedes the necessity of involving the ratio in each particular calculation.

These powers show the Amount of $1 for each corresponding rate and number of years.

Thus the Amount of $1, at 5 per cent, for 1 year, is $1.05; for 2 years it is $1X(1.05)=(1.05)2; for 3 years, $1X(1.05)3 =(1.05)3; and so on, in the ascending powers of the ratio 1.05, (§ 357).

Hence the following methods

§ 359. 1. To find the Amount at Compound Interest,-Multiply the Principal by tabular amount of $1 for the given råte and number of years or periods; and

2. To find the Principal,-Divide the given Amount by the tabular amount of $1 for the given rate and number of years or periods for which interest is computed.

Thus to find what $100 would amount to in 20 years, at 6 per cent, allowing the interest to be compounded annually, Opposite to 20 years, and under 6 per cent, in the TABLE, we find that the amount of $1 would be $3 20713'.

Then $3.20713X100 Answer $320.713'.

When the Number of Years or Periods exceeds 50.

§ 360. The Amount of $1 at Compound Interest for any number of years or periods, multiplied by the Amount for any other number, produces the Amount of $1 for the sum of those two numbers, (§ 295.)

Thus 18.42015, the Amount of $1, for 50 years, at 6 per cent, X 4.29187', the amount for 25 years, 79.05688', the Amount of $1, at 6 per cent, for 75 years.