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2. The first term of a geometrical progression is 1, the last

term is 2048, and the number of terms is 12.

sum of all the terms?

What is the

Ans. 4095.

3. The first term of a geometrical progression is 1, the last term is 19683, and the number of terms is 10.

sum of all the terms?

CASE XII.

What is the

Given the ratio, the number of terms, and the last term, to find the sum of all the terms.

RULE.

Raise the ratio to a power, whose exponent is the number of terms, from this power subtract one, and multiply the remainder by the last term; divide this product by the product of the ratio, less one, into the power of the ratio, whose exponent is the number of terms, less one.

Examples.

1. The ratio of the terms of a geometrical progression is 2, the number of terms is 12, and the last term is 2048. What is the sum of all the terms?

In this example the ratio, raised to a power, whose exponent is the number of terms, is 212=4096, this, diminished by one, becomes 4095, which, multiplied by 2048, becomes 8386560; again, the power of the ratio, whose exponent is one less than the number of terms, is 2048, which, multiplied by the ratio, less one, is not changed; ... 8386560 divided by 2048, gives 4095, for the sum of all the terms.

2. The ratio of the terms of a geometrical progression is 3, the number of terms is 8, and the last term is 106. What is the sum of all the terms? Ans. 307.

3. The ratio of the terms of a geometrical progression is , the number of terms is 7, and the last term is 2581877. What is the sum of all the terms?

NOTE.-The eight remaining cases in geometrical progression, can not be solved by the ordinary processes of arithmetic, but require for their solution, a knowledge of logarithms and algebraic equations, above the second degree.

64. When the ratio of a geometrical progression is less than a unit, the first term will be the largest, and the last term the least-the progression will, in this case, be descending; but if we consider the series of terms in a reverse order, that is, calling the last term the first, and the first the last, the progression may then be considered as ascending.

If a decreasing geometrical progression be continued to an infinite number of terms, we may neglect the last term as of no appreciable value; we can find its sum by case II., when it is modified, as follows:

Given the first term of a descending geometrical progression, and the ratio, to find the sum of all the terms, when continued to infinity.

RULE.

Divide the first term by a unit diminished by the ratio.

Examples.

1. What is the sum of all the terms of the infinite series 1, 1, 1, 1, &c.?

In this example a unit, diminished by the ratio, is 1-}=}, and the first term, 1, divided by 1, gives 2, for the sum of all the terms.

2. What is the sum of the infinite series 1,,, 44, &c.? Ans. 11.

3. What is the sum of the infinite series, 1, 1000, TOUT, &c.?

37

Ans. 4.

4. What is the sum of the infinite series. I†õ, Tõõ0, TOOT, &c.?

16

Ans. 1.

5. What is the sum of the infinite series 180, Tõ‰õõ, Ans. TOO BOOT, &c.?

6. What is the sum of the infinite series, 180, 1800,

[blocks in formation]

63

7. What is the sum of the infinite series 10, 10000, TOO700, &c.?

567

From a mere inspection of the above examples, we discover that a vulgar fraction whose numerator is 1, when converted into a decimal fraction, forms a geometrical series, whose first term agrees with the first significant figure in the decimal; the ratio is the first remainder which was obtained when dividing by the denominator.

CHAPTER VIII.

SIMPLE INTEREST.

65. Interest is money paid by the borrower to the lender, for the use of the money borrowed.

It is estimated at a certain per cent. per annum, that is, a certain number of dollars for the use of $100, for one year.

Thus, when $6 is paid for the use of $100, for one year, the interest is said to be at 6 per cent.

In the same manner when $5 is paid for the use of $100, for one year, the interest is said to be at 5 per cent.; and the same for other rates.

The rate per cent. is generally fixed by law. In the New England States the legal rate is 6 per cent., whilst in the State of New York it is 7 per cent.

The sum of money borrowed, or upon which the interest is computed, is called the principal.

The principal, with the interest added to it, is called the

amount.

CASE I.

To find the interest on $1, for any given time, at 6 per cent. The interest on $100, for one year, at 6 per cent., being $6, it follows that the interest on $1, for one year, is $0.06; and since 2 months is of a year, the interest on $1, for 2 months, is $0.01; again, since 6 days is of 2 months, when we reckon 30 days to each month, it follows that the

=

interest on $1, for 6 days, is $0.001. Hence, we have the following

RULE.

Call half the number of months, CENTS; one sixth the number of days, MILLS.

Examples.

1. What is the interest of $1, for 7 months and 10 days, at 6 per cent.?

In this example half the number of months is 31, which being called cents, gives $0.035, for the interest of $1, for 7 months; again, one-sixth of the number of days is 1, which, being called mills, gives $0.001, for the interest of $1, for 10 days; therefore, the interest for $1, for 7 months and 10 days, is $0.0361.

2. What is the interest of $1, for 11 months and 11 days, at 6 per cent.? Ans. $0.056.

3. What is the interest of $1, for 3 years 7 months, that is, for 43 months, at 6 per cent.? Ans. $0.215.

4. What is the interest of $1, for 2 years 7 months and 9 days, at 6 per cent.? Ans. $0.1565.

5. What is the interest of $1, for 1 year 7 months and 15 days, at 6 per cent.? Ans. $0.0975. 6. What is the interest of $1, for 7 years and 9 days, at 6 Ans. $0.4215. per cent.?

7. What is the interest of $1, for 3 years 5 months and 3 days, at 6 per cent.? Ans. $0.2055.

8. What is the interest of $1, for 9 years and 3 months, at 6 per cent.?

9. What is the interest of $1, för 21 years 5 months and 6 days, at 6 per cent.?

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