EXAMPLES.

1. What is the 100th term of an arithmetical progression, whose first term is 2, and common difference 3?

In this example, the number of terms, less one, is 99, which, multiplied by the common difference, 3, gives 297, which, added to the first term, 2, makes 299 for the 100th term.

2. What is the 50th term of the arithmetical progression, whose first term is 1, the common difference ? Ans. 251.

3. A man buys 10 sheep, giving $1 for the first, $3 for the second, $5 for the third, and so on, increasing in arithmetical progression. What did the last sheep cost at that rate? Ans. $19.

4. The first term of an arithmetical progression is the common difference, and the number of terms 26. What is the last term? Ans. 37.

CASE II.

From the nature of an arithmetical progression, we see that the second term added to the next to the last term is equal to the first added to the last; since the second term is as much greater than the first, as the next to the last is less than the last. After the same method of reasoning, we infer that the sum of any two terms equidistant from the extremes, is equal to the sum of the extremes.

Hence, it follows that the terms will average just half the sum of the extremes.

Therefore, when we have given the first term, the last term, and the number of terms, to find the sum of all the terms, we have this

RULE.

Multiply half the sum of the extremes by the number of

terms.

EXAMPLES.

1. The first term of an arithmetical progression is 2, the last term is 50, and the number of terms is 17. What is the sum of all the terms?

In this example, half the sum of the extremes is

of (2+50) 26.

This, multiplied by the number of terms, gives 26 × 17= 442, for the sum required.

What

2. The first term of an arithmetical progression is 13, the last term is 1003, the number of terms is 100. is the sum of the progression?

Ans. 50800.

3. A person travels 25 days, going 11 miles the first day, 135 the last day; the miles which he traveled in the successive days, form an arithmetical progression. How far did he go in the 25 days? Ans. 1825 miles.

4. Bought 7 books, the prices of which are in arithmetical progression. The price of the first was 8 shillings, and the price of the last was 28 shillings. What did they all come to? Ans. £6 6s.

5. What is the sum of 1000 terms of an arithmetical progression, whose first term is 7 and last term 1113? Ans. 560000.

6. The first term of an arithmetical progression is 4, and the last term 3654, and the number of terms 799. What is the sum of all the terms?

Ans. 146217.

GEOMETRICAL PROGRESSION.

138. A SERIES of numbers which succeed each other regularly, by a constant multiplier, is called a geometrical progression.

This constant factor, by which the successive terms are multiplied, is called the ratio.

When the ratio is greater than a unit, the series is called an ascending geometrical progression.

When the ratio is less than a unit, the series is called a descending geometrical progression.

Thus, 1, 3, 9, 27, 81, &c., is an ascending geometrical progression, whose ratio is 3.

And, 1, 4, &c., is a descending geometrical progression, whose ratio is .

In geometrical progression, as in arithmetical progression, there are five things to be considered.

1 The first term.

2. The last term.

3. The common ratio.

4. The number of terms.

5. The sum of all the terms.

These quantities are so related to each other, that any three being given, the remaining two can be found.

The solution of some of these cases requires a knowledge of higher principles of mathematics than can be detailed by arithmetic alone.

We will give a demonstration of the rules of one or two of the most important cases.

When are numbers in geometrical progression? What is the constant factor, by which the successive terms are multiplied, called ? When this ratio exceeds a unit, the progression is called what? When this ratio is less than a unit, how is the progression called? Give an example of an ascending geometrical progression? Give an example of a descending geometrical progression. How many quantities are to be considered in geometrical progression? Mention these quantities. How many of these must be known to enable us to find the others ?

CASE I.

By the definition of a geometrical progression, it follows that the second term is equal to the first term, multiplied by the ratio; the third term is equal to the first term, multiplied by the second power of the ratio; the fourth term is equal to the first term, multiplied by the third power of the ratio; and so on, for the succeeding terms.

Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this

RULE.

Multiply the first term by the power of the ratio, whose exponent is one less than the number of terms.

EXAMPLES.

1. The first term of a geometrical progression is 1, the ratio is 2, and the number of terms is 7. What is the last term?

In this example, the power of the ratio, whose exponent is one less than the number of terms, is 2o=64, which, multiplied by the first term, 1, still remains 64, for the last

term.

2. The first term of a geometrical progression is 5, the ratio is 4, and the number of terms 9. What is the last

term?

27

Ans. 327680.

3. A person traveling, goes 5 miles the first day, 10 miles the second day, 20 miles the third day, and so on, increasing in geometrical progression. If he continue to travel in this way for 7 days, how far will he go the last day? Ans. 320 miles.

CASE II.

Let it be required to find the sum of all the terms of the geometrical progression 2, 6, 18, 54, 162, 486,

If we multiply each term by 3, which is the ratio, we shall obtain this second progression, 6, 18, 54, 162, 486, 1458, the sum of whose terms is obviously 3 times as great as the sum of the terms of the first progression. Consequently, the difference between the sums of the terms of these two progressions is (3-1)=2 times the sum of all the terms of the first progression. If we omit the first term of the first progression, it will agree with the second progression, after omitting its last term. Hence, the difference between the sums of the terms of these two progressions may be found by subtracting 2, the first term of the first progression, from 1458, the last term of the second progression; but 1458 was obtained by multiplying 486, the last term of the first progression, by 3, the ratio.

Hence, we finally obtain this condition:

That the sum of all the terms of a geometrical progression, repeated as many times as there are units in the ratio, less one, is equal to the last term multiplied by the ratio, and diminished by the first term.

Therefore, when we have given the first term of a geo