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d.

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3. The number of terms, represented by n.
4. The common difference,

5. The sum of all the terms, 199. The two fundamental formulas of arithmetical progression are:

I. The formula for the last term, 1 = a + (n 1)d.
This formula is derived from a series as follows:

(1) (2) (3) (4) (5) (6)

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, I = a + (n-1)d.

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

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

In a descending series, d is subtracted and the formula is written l = a – (n − 1)d. II. The formula for the sum of the series, S= (a+1).

2 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,

s = a + (a + d) + (a + 2d) + (l – 2d) + (l d) + l. or, s= l + (1 d) + (1 2d) + (a + 2d) + (a + d) + a. Adding, 2s=(a +1) +(a +1)+(a +1) + . . (a +1)+(a+1)+(a +1). ..28 = (a + l) taken as many times as there are terms (n).

(a + 1) .. 2s = n(a + 1), and s = (a + 1) = n

2

n

n

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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, SOLUTION: 1= a + (n 1)d; ..1 2 + (12 – 1) 2 = 24, ans. 2. Find the 14th term of the series 64, 61, 58, 55, .... SOLUTION: 1

(n 1)d; .l=64 – (14 – 1)3 = 25, ans. 3. Find the 1st term of the series . ... 68, 71, 74, having 19 terms.

SOLUTION: 1=a + (n − 1)d; .:.74 = a + (19 - 1)3; a=20, ans.

= a

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?

SOLUTION: 1 = a + (n -1)d; . '. 23 = 2 + (n − 1)3, and n=8,

ans.

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

12 SOLUTION: S= (a + D; :'.8= (5 + 32). = 222, ans. 2

2

Exercise XXII

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

2. Find the 1st term of the series. 53, 56, 59, 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, 1 = 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, Since 1 = a + (n 1)d, s =

2

[a + a + (n 1)d].

n

Substituting in the 2d formula, we have

(1) 787.92

n
2

[16.08 + 16.08 + (n − 1) 32.16]

n (2) 787.92

2

(32.16 + 32.16 n 32.16)
(3) 787.92 = 16.08 n?
(4) 49 = n2

7 =

= n;.. 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.
2. The last term,

l.

r.

S.

.

3. The number of terms, represented by n.
4. The ratio,

5. The sum of all the terms, 204. The two fundamental formulas of geometrical progression are:

I. The formula for the last term, 1 = aron - 1. This formula is derived from a series as follows: (1) (2) (3) (4) (5) (6) (7) (n). a, ar, ar, ars, ar, ars,

1. 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 arh-1; or, letting I stand for the nth term, or last term, 1 = ar-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.

are,

II. The formula for the sum of the series,

arn

a S =

r 1 This formula may be derived as follows:

s= a + ar + ar? + am? + .. ar» – 2 + ar-1. rs = ar + ara + aml + ar—2 + arn-1 + ars.

s= arn s (r - 1) = arn aron a (pon – 1)

Ir - a r - 1

-1' 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.

rs

a.

a.

- a

S =

or s

or s

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-1

r

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