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any two terms equally distant from the extremes, or to twice the middle term, when the number of terms is odd.

1, 5, 9, 13, 17, 21, 25, 29, 33, 37. 37, 33, 29, 25, 21, 17, 13, 9, 5, 1.

38, 38, 38, 38, 38, 38, 38, 38, 38, 38.

We thus have 10 times 38, or the number of terms multiplied by the sum of the extremes, which is equal to twice the sum desired. Hence we derive the following

RULE.

To obtain the last term, multiply the common difference by the number of terms less one, and if the series is ascending, add the product to the first term,-if decreasing, subtract it from the first term.

To obtain the sum of the terms, multiply the number of terms by the sum of the extremes, and divide the product by two.*

1. The first term of an equidifferent series is 5, the common difference 2, and the number of terms 11. Find the last term, and the sum of all the terms.

2. What is the sum of 45 terms of the natural series, 1, 2, 3, &c.?

3. Find the 23d term, and the sum of 23 terms of the series 1, 4, 7, 10, &c.

4. How many times does a clock strike in 12 hours?

5. The first term of a descending series is 30, the common difference, and the number of terms 33. the last term, and the sum of the series.

Required

6. If I travel 100 miles to-day, and 5 miles less on each succeeding day, how far shall I have gone at the end of the 17th day, and what will be the length of the last day's journey?

PROBLEM II.

The extremes and number of terms being given, to find the common difference.

The first term of an equidifferent series is 27, the last term 3, and the number of terms 9. What is the common difference?

The last term is found by Problem I. by subtracting 8 times

*If the given term is the last term, invert the series, and work by the same rule.

Then the difference

the common difference from the first term. of the extremes 24, must be 8 times the common difference, which is therefore 3.

RULE.

Divide the difference of the extremes by the number of terms less one, and the quotient will be the common difference. This difference repeatedly added to the less, or subtracted from the greater term, will give the intermediate terms.

The sum of the terms is found by the preceding problem. 7. One hundred and fifty eggs were laid in a row. The first was 2 yards, and the last 100 yards from a basket. How far were the eggs apart, and how far must a person travel to pick them up singly and lay them in the basket?

8. A man had 15 children, whose ages were in Arithmetical Progression: the youngest was 3 years old, and the oldest 24. Required the common difference of their ages.

9. The extremes of an equidifferent series are 1 and 11, and the number of terms 25. What are the mean terms?

10. Insert 12 arithmetical means between 20 and 59. [As there are 12 means, the number of terms must be 14.] 11. Insert 13 arithmetical means between 3 and 4.

12. A man commenced walking for exercise, daily increasing the distance by equal additions. The first day he walked 4 miles, and on the 15th day he walked 7 miles. What was the daily increase?

PROBLEM III.

The extremes and common difference being given, to find the number of terms?

The extremes are 45 and 10, and the common difference is 5. What is the number of terms.

By Problem I, the difference of the extremes is equal to the common difference multiplied by the number of terms less one. Then the difference of the extremes 35, divided by the common difference 5, gives 7, which must be equal to the number of terms less 1. Therefore the number of terms is 8.

RULE.

Divide the difference of the extremes by the common difference, and add 1 to the quotient.

13. What is the sum of the series 2, 4, 6, 8, &c., to 1000?

14. What is the sum of the descending series 100, 99, 99, 98, &c., the last term being 0?

15. The first term of a series is 11, the last term 2, and the common difference. What is the number of terms?

16. The first term 24, the common difference, and the last term 90, being given, required the number of terms.

PROBLEM IV.

The extremes and sum of the terms being given, to find the number of terms.

RULE.

Divide twice the sum of the terms by the sum of the

extremes.

17. The sum of the terms is 221, the first term 4, and the last term 11. Required the number of terms and the common difference.

PROBLEM V.

One of the extremes, the number of terms, and the sum of the terms being given, to find the other extreme.

RULE.

Divide twice the sum of the terms by the number of the terms, and subtract the given extreme from the quotient.

18. The sum of the terms in an equidifferent series is 27, one of the extremes is 4, and the number of terms 8. Required the other extreme and the common difference. PROBLEM VI.

The number of terms, common difference, and sum of the terms being given, to find the extremes.

RULE.

Divide the sum of the terms by the number of terms. Then multiply the common difference by the number of terms less one, and subtract half the product fron the first quotient for one extreme. For the other extreme add half the product to the quotient. Problems IV., V., and VI., are all readily deduced from Pro

blem I.

CHAPTER XVIII.

GEOMETRICAL PROGRESSION, OR CONTINUAL PROPORTIONALS.

A SERIES of numbers whose successive terms increase or diminish uniformly in the same ratio, is called a PROGRESSION BY QUOTIENT, or GEOMETRICAL PROGRESSION. The numbers may also be regarded as a series of CONTINUAL PROPORTIONALS. The ratio has already been defined in the Chapter on Proportion. As in Arithmetical Progression, the first and last terms are called the extremes, the other terms the means.

Thus in the ascending progression,

1, 3, 9, 27, 81, 243, the extremes are 1 and 243, and the ratio is 3. In the descending progression,

3 3

24, 12, 6, 3, 3, 3, 4, 7, the extremes are 24 and, and the ratio is

From the nature of the series, it is evident that any four successive terms of a Geometrical Progression, constitute a proportion; as 1 : 3 9:27, 3:9:: 27: 81, &c., in the first of the above series; 12: 6:3:3, 6:3, &c. in the second series.

PROBLEM I.

One of the extremes, the ratio, and the number of terms being given, to find the other extreme.

The first term of an increasing geometrical series is 2, and the ratio 3. What is the sixth term?

The second term will be 2 X 3; the third, 2 x 3 x 3 or 2 × 32; the fourth, 2 x 32 × 3, or 2 × 33, and so on, to the sixth, which is 2 x 35. If the sixth term had been given, and the first required, we should evidently have been obliged to divide by 35.

RULE.

Raise the ratio to a power whose index is equal to the number of terms less one. Then for the last term multiply, and for the first term divide, the given extreme by this power of the ratio.

We have already seen in Involution, that by adding the exponents of two powers of the same number, we shall obtain the exponent of their product. Thus 37 x 34 311; 52 X 53 55, &c. This princi ple will greatly assist us in finding the powers of the ratio.

What is the 17th power of 2?

1 2 3 4

2, 4, 8, 16. 65536 216

24 X 24 = 256 28
216 X 2 131072 217.

=

28 X 28 256x256=

=

In this instance we form the powers as high as the 4th power, then multiply the 4th power by itself for the 8th power,-the 8th power by itself for the 16th power, and the 16th power by the 1st power for the 17th power. We have, therefore, after obtaining the 4th power, only three multiplications to make, instead of thirteen, which would otherwise have been necessary.

1. The first term of a geometrical series is 3, the ratio 2, and the number of terms 9. What is the last term?

3. What is the 11th term of the series 4096, 1024, 256, &c.?

3. What is the 7th term of a series, whose first term is 20, and ratio 1.06 ?

4. What is the amount of $20 for 7 years, at 6 per cent. compound interest?

5. What is the amount of $300 for 9 years, at 6 per cent. compound interest?

6. If the first pane of glass in a window cost 1 mill, the second 2 mills, the third 4 mills, the 4th 8 mills, &c., what would be the price of the twelfth pane?

7. If $100 were at simple interest for three years at 6 per cent., and the amount then placed at compound interest at 5 per cent., what would be the whole amount at the expiration of 12 years?

8. The first term is 8, the ratio, and the number of terms 8. What is the last term?

9. The twelfth term is 59049, and the ratio. What is the first term?

10. What principal will amount to $4489.643, in 12 years, at 5 per cent. compound interest? +

11. If a farmer plants a grain of wheat, and each year plants the product of the preceding harvest, how much will

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