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By considering the manner in which the ascending progression is formed, we see that the 2d term is obtained by adding the common difference to the first term; the 3d by adding the common difference to the 2d; the 4th by adding the common difference to the 3d, and so on; the number of additions being 1 less than the number of the term found.

But instead of making the additions, we may multiply the common difference by the number of additions, that is by 1 less than the number of terms, and add the first term to the product. Hence we have

CASE I. $ 225. Having given the first term, the common difference, and the number of terms, to find the last term.

RULE. Multiply the common difference by 1 less than the number of terms, and to the product add the first term.

Ex. 1. The first term 3, the common difference 2, and the number of terms 19: what is the last term ?

18 number of terms less 1.

2 common difference 36

3 1st term 39 last term.

Ans. 39. 2. A man bought 50 yards of cloth for which he was to pay 6 cents for the first yard, 9 cents for the 2d, 12 cents for the 3d, and so on increasing by the common difference 3: how much did he

pay

for the last yard ?

Ans. $1,5

3. A man puts out $100 at simple interest, at 7 per cent. ; at the end of the first year it will have increased to $107, at the end of the 2d year to $114 and so on, increasing $7 each year: what will be the amount at the end of 16 years ?

Ans. $205. Since the last term of an arithmetical progression is equal to the first term added to the product of the common difference by 1 less than the number of terms, it follows, that the difference of the extremes will be equal to this product, and that the common difference will be equal to this product divided by 1 less than the number of terms. Hence we have

CASE II. § 226. Having given the two extremes and the number of terms of an arithmetical progression, to find the common difference.

RULE: Subtract the less extreme from the greater and divide the remainder by 1 less than the number of terms, the quotient will be the common difference.

Ex. 1. The extremes are 4 and 104, and the num. ber of terms 26 : what is the common difference?

104

4 26-1=25)100(4 100

Ans. 4. 2. A man has 8 sons, the youngest is 4 years old and the eldest 32, their ages increase in arithmetical progression : what is the common difference of their

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ages ?

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Ans. 4.

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3. A man is to travel from New York to a certain place in 12 days, and to go 3 miles the first day, 1 increasing every day by the same number of miles,

so that the last day's journey may be 58 miles : required the daily increase.

Ans. 5 miles. If we take any arithmetical series, as 3 5 7 9 11 13 15 17 19, &c.

19 17 15 13 11 9 5 3 by reversing the order € 22 22 22 22 22 22 22 22 22 of the terms.

Here we see that the sum of the terms of these two series is equal to 22, the sum of the extremes, multiplied by the number of terms; and consequently, the sum of either series is equal to the sum of the two extremes multiplied by half the number of terms. Hence we have

CASE III.

$ 227. To find the sum of all the terms of an arithmetical progression.

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RULE.

Add the two extremes together and multiply their sum by half the number of terms, the product will be the sum of the series,

Ex. 1. The extremes are 2 and 100, and the number of terms 22: what is the sum of the series?

2 1st term
100 last term
102 sum of extremes

11 half the number of terms
1122 sum of series.

Ans. 1122.

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2. How many strokes does the hammer of a clock strike in 12 hours ?

Ans. 78. 3. If 100 eggs were placed in a right line, exactly one yard from each other, and the first, one yard from a basket: what distance will a man travel who gathers them up singly, and places them in the basket ?

Ans. 5 miles, 1300 yards.

GEOMETRICAL PROGRESSION.

§ 228. If we take any number, as 3, and multiply it continually by any other number, as 2, we form & series of numbers, thus,

3 6 12 24 48 96 192, &c. in which each number is formed by multiplying the number before it by 2.

This series may also be formed by dividing continually the largest number 192 by 2. Thus,

192 96 48 24 12 6 3 A series formed in either

way

is called a Geome. trical Series, or a Geometrical Progression, and the number by which we continually multiply or divide, is called the common ratio.

When the series is formed by multiplying continually by the common ratio, it is called an ascendo ing series ; and when it is formed by dividing con. tinually by the common ratio, it is called a descending series.

Thus,
3 6 12 24

48 96 192 is an ascending series. 192 96 48 24 12

6 3 is a descending series

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The several numbers are called the terms of the progression.

The first and last terms are called the extremes, and the intermediate terms are called the means.

$ 229. In every Geometrical, as well as in every
Arithmetical Progression, there are five things which
are considered, any three of which being given or
known, the remaining two can be determined.
They are,

1st the first term,
2d the last term,
3d the common difference,
4th the number of terms,

5th the sum of all the terms.
By considering the manner in which the ascende
ing progression is formed, we see that the second
term is obtained by multiplying the first term by the
common ratio; the 3d term by multiplying this pro-
duct by the common ratio, and so on, the number of
multiplications being one less than the number of
terms. Thus,

3=1 lst term, 3x2=6 2d term, 3x2x2=12 3d term, 3x2x2x2=24 4th term, &c. for the other terms. But 2x2=2", 2x2x2=2', and 2x2x2x2=2.

Therefore, any term of a progression is equal to the first term multiplied by the ratio raised to a power 1 less than the number of the term.

CASE I.

§ 230. Having given the first term, the common ratio, and the number of terms, to find the last term

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