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3. The number of terms, represented by n.

4. The common difference,

5. The sum of all the terms,

66

“d.

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66

S.

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This formula is derived from a series as follows:

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a, a+d, a+2d, a+3d, a+4d, a+5d, etc.

From the series we observe that any term consists of the first term the common difference taken as many times as there are terms less 1. Hence, for the nth term we have a + (n − 1)d; or, letting I stand for the nth term, or last term, la + (n-1)d.

=

Study the formula and then write the rule for finding the last term in an arithmetical series.

Write

From the above formula, derive the formula for finding the first term; the number of terms; the common difference. the rule corresponding to each formula.

In a descending series, d is subtracted and the formula is written l = α (n - 1)d.

II. The formula for the sum of the series, s=(a+1).

This formula may be derived as follows:

Since the sum of a series is simply the sum of all the terms in the series, we may write,

8 = a + (a + d) + (a + 2d) + or, 8 1 + (1 − d) + (1 − 2d) +

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(1 − 2d) + (l

d) + l.

(a + 2d) + (a + d) + a.

Adding, 28= (a + 1) + (a + 1) + (a + 1) + . . .. (a + 1) + (a+1)+(a + 1). ... 28 (a + 1) taken as many times as there are terms (n).

n

... 28

n(a + 1), and s

= 2/2 (a + 1) = n

(a + 1)
2

Study this formula, and then write the rule for finding the sum of an arithmetical series. From the formula above, derive the formula for finding the first term; the number of terms; the last term. Write the rule corresponding to each formula.

EXAMPLES

1. Find the 12th term of the series 2, 4, 6, 8, . . . .

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2. Find the 14th term of the series 64, 61, 58, 55,

SOLUTION: 1 = a - (n − 1)d; ... 7 =

64

3. Find the 1st term of the series

having 19 terms.

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(141)3 = 25, ans.

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SOLUTION: 1= a + (n − 1)d;

.74 =

= a + (19 − 1)3; a=20, ans.

4. If the extremes of an arithmetical series are 3 and 15, and the number of terms 7, what is the common difference? SOLUTION: 1 = a + (n − 1)d; . '. 15 = 3 + (7 − 1)d, and d = 2,

ans.

5. If the extremes are 2 and 23, and the common difference 3, what is the number of terms?

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6. If the extremes are 5 and 32, and the number of terms 12, what is the sum of the series?

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1. Find the 9th term of the series 3, 5, 7,

2. Find the 1st term of the series having 10 terms.

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3. The extremes are 1 and 51, and the number of terms is 76. What is the common difference?

4. If a is 2, and I is 17, and n is 6, what is d?

5. If the extremes are 5 and 75, and the common difference 5, what is the number of terms?

6. How many strokes does a clock make in 12 hours?

7. 40 potatoes are 2 yd. apart and the first is 2 yd. from a basket. How far will a boy travel who gathers them and puts them into the basket one at a time?

8. Find s; given a = 10, n = 6, d = 4. Write out the series.

9. Find n; given s = 36160, a = 40, l = 600.

10. What is the sum of the first fifty odd numbers?

11. What is the sum of the first 50 numbers divisible by 7?

12. A body, if left unsupported, will fall by its own weight during the 1st second, 16.08 ft. (N. Y.); in each succeeding second, it will fall 32.16 ft. farther than in the preceding second; how far will it fall in the 11th second? In 11 seconds?

13. A boy throws a stone into the air, and it strikes the ground in 4 sec. How high did the stone go?

14. A bag of sand dropped from a balloon and fell to the earth in a quarter of a minute. How high was the balloon?

15. A body passed over 787.92 feet during its fall; what was the time required?

Remember that d = 32.16. Combine the two formulas for problems similar to the 15th; thus,

n

Since l = a + (n − 1)d, s =

2

[a + a + (n − 1)d].

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(5)

49

n2

7n; .. the time was 7 seconds.

16. A stone is thrown horizontally from the top of a tower 257.28 ft. high, with a velocity of 60 ft. a second. Where will it strike the ground?

17. A body falls freely for 6 seconds. What is the space traversed during the last 2 seconds of its fall?

18. A body falls from a certain height; 3 seconds after it has started, another body falls from the height of 787.92 feet; from what height must the first fall, if both are to reach the ground at the same instant?

II. GEOMETRICAL PROGRESSION

200. A geometrical progression is a series of numbers increasing or decreasing by a constant multiplier.

201. If the multiplier is greater than a unit, the series is ascending. If the multiplier is less than a unit, the series is descending.

202. The ratio is the constant multiplier.

203. In a geometrical progression there are five parts, any three of which being given, the other two can be found. The five parts with which we are concerned are:

1. The first term, represented by a.

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3. The number of terms, represented by n.

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This formula is derived from a series as follows:

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From the series we observe that any term consists of the first term multiplied by the ratio raised to the power indicated by a number which is 1 less than the number of the term. Hence, for the nth term, we have ar"-1; or, letting I stand for the nth term, or last term, l arn-1.

=

Study the formula and then write the rule for finding the last term in a geometrical series.

From the above formula, derive the formula for finding the first term; the ratio. Write the rule corresponding to each

formula.

II. The formula for the sum of the series,

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This formula may be derived as follows:

s = a + ar + ar2 + ar? + ... ar2 -2 + ɑrn−1.

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Study this formula, and then

write the rule for finding the sum of the geometrical series. From the formula above, derive the formula for finding the first term; the last term; the ratio. Write the rule corresponding to each formula.

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